From Functions to Equivalence Classes
Let \((\Omega, \mathcal{F}, \mu)\) be a measure space. In
Intro to Functional Analysis, we
introduced the \(L^p\) space as the collection of measurable functions \(f\) with
\(\int |f|^p \, d\mu \lt \infty\), taken modulo equality almost everywhere, equipped with the
quantity
\[
\|f\|_p = \left( \int_\Omega |f|^p \, d\mu \right)^{1/p}.
\]
For this to be a valid norm, three axioms must hold:
- (i) positive definiteness:
\(\|f\|_p \geq 0\), with \(\|f\|_p = 0 \iff f = 0\). The "\(\Leftarrow\)" direction is immediate, and the
"\(\Rightarrow\)" direction is the non-trivial content.
- (ii) absolute homogeneity:
\(\|\alpha f\|_p = |\alpha| \cdot \|f\|_p\).
- (iii) triangle inequality:
\(\|f + g\|_p \leq \|f\|_p + \|g\|_p\).
Non-negativity and axiom (ii) are straightforward: the former from the non-negativity of the integrand
\(|f|^p\), the latter from the linearity of the integral. The two remaining difficulties are the
definiteness half of axiom (i) and the triangle inequality. Together they drive the structure of
this entire page. We address definiteness first.
Convention. Throughout this page, \(\mathbb{F}\) denotes the scalar field, either
\(\mathbb{R}\) or \(\mathbb{C}\). Measurable functions \(f : \Omega \to \mathbb{F}\) are understood as
\((\mathcal{F}, \mathcal{B}(\mathbb{F}))\)-measurable, where \(\mathcal{B}(\mathbb{F})\) is the Borel
\(\sigma\)-algebra on \(\mathbb{F}\). Results are stated in generality over \(\mathbb{F}\). Where the
real and complex cases need separate treatment, the text says so at that point. Such points are of
two kinds: arguments run on real and imaginary parts separately, and phase factors. The latter
appear in the equality condition for Minkowski and in the extremal functions built for the
duality bound.
The Seminorm Problem
Suppose \(\|f\|_p = 0\). Then \(\int |f|^p \, d\mu = 0\). Since \(f\) is measurable, so is the
composition \(|f|^p\) (the absolute value is continuous on \(\mathbb{F}\) and \(t \mapsto t^p\) is
continuous on \([0, \infty)\), both are Borel measurable, and measurability is preserved under
composition), and \(|f|^p \geq 0\) pointwise. By
Theorem: Zero Integral Implies Vanishing Almost Everywhere,
\(|f(x)|^p = 0\) for almost every \(x \in \Omega\). Since \(p \geq 1\), the map \(t \mapsto t^p\) is
strictly increasing on \([0, \infty)\), so \(t^p = 0 \iff t = 0\). Hence \(|f(x)| = 0\) a.e., that is,
\(f(x) = 0\) except on a set of measure zero.
But this does not mean \(f\) is the zero function. It means only that \(f = 0\) almost
everywhere (a.e.).
For a concrete example, recall the Lebesgue Integration
of the
Dirichlet function:
\[
f(x) = \chi_{\mathbb{Q}}(x) =
\begin{cases}
1 & \text{if } x \in \mathbb{Q}, \\\\
0 & \text{if } x \notin \mathbb{Q}.
\end{cases}
\]
Considering \(\mathbb{R}\) with Lebesgue measure, since \(\mu(\mathbb{Q}) = 0\), we have \(\|f\|_p = 0\)
for every \(1 \leq p \lt \infty\), yet \(f\) is not the zero function. It equals \(1\) at every rational
point, so the quantity \(\|\cdot\|_p\) fails to distinguish \(f\) from the zero function. In the language
of normed space theory, \(\|\cdot\|_p\) is a seminorm, not a norm. It satisfies
\(\|f\|_p = 0\) without \(f = 0\).
The Equivalence Relation
The resolution is a standard algebraic maneuver. We quotient out the ambiguity by declaring two
measurable functions to be "the same" when they differ only on a negligible set.
Definition: Equality Almost Everywhere
Let \(f, g : \Omega \to \mathbb{F}\) (\(\mathbb{F} = \mathbb{R}\) or \(\mathbb{C}\))
be measurable functions. We say \(f\) and \(g\) are
equal almost everywhere, written \(f = g\) a.e., if
\[
\mu\bigl(\{x \in \Omega : f(x) \neq g(x)\}\bigr) = 0.
\]
The relation \(f \sim g \iff f = g\) a.e. is an equivalence relation on
the set of measurable functions.
Verification:
We verify the three axioms of an equivalence relation, beginning with a preliminary remark. For
measurable \(f, g : \Omega \to \mathbb{F}\), the difference \(f - g\) is measurable (componentwise in
the complex case). Thus \(\{x : f(x) = g(x)\} = (f - g)^{-1}(\{0\})\). Since \(f - g\) is measurable
(that is, the preimage of every Borel set lies in \(\mathcal{F}\)) and \(\{0\} \subset \mathbb{F}\)
is closed hence Borel, this preimage is in \(\mathcal{F}\), and so is its complement
\(\{x : f(x) \neq g(x)\}\).
Reflexivity. We have \(\{x : f(x) \neq f(x)\} = \emptyset\), which has measure zero.
Symmetry. The sets \(\{f \neq g\}\) and \(\{g \neq f\}\) are literally equal, so
\(f \sim g \iff g \sim f\).
Transitivity. Suppose \(f = g\) a.e. and \(g = h\) a.e. Let
\(N_1 = \{x : f(x) \neq g(x)\}\) and \(N_2 = \{x : g(x) \neq h(x)\}\). Both are null by hypothesis
and measurable by the preliminary remark (applied to each pair). The set \(\{x : f(x) \neq h(x)\}\)
is likewise measurable by the remark (applied to \(f\) and \(h\)), and if \(f(x) \neq h(x)\), then
either \(f(x) \neq g(x)\) or \(g(x) \neq h(x)\), so
\(\{x : f(x) \neq h(x)\} \subseteq N_1 \cup N_2\). By monotonicity and subadditivity of \(\mu\),
\(\mu(\{x : f(x) \neq h(x)\}) \leq \mu(N_1) + \mu(N_2) = 0\).
The Formal Definition of \(L^p\)
Definition: The \(L^p\) Space (Rigorous)
Let \((\Omega, \mathcal{F}, \mu)\) be a measure space and \(1 \leq p \lt \infty\). Define
\[
\mathscr{L}^p(\Omega, \mathcal{F}, \mu) =
\bigl\{\, f : \Omega \to \mathbb{F} \mid f \text{ is measurable and }
\int_\Omega |f|^p \, d\mu \lt \infty \,\bigr\}.
\]
The \(L^p\) space is the quotient
\[
L^p(\Omega, \mathcal{F}, \mu) = \mathscr{L}^p(\Omega, \mathcal{F}, \mu) \,\big/\!\sim
\]
where \(f \sim g \iff f = g\) a.e. Each element of \(L^p\) is an equivalence class
\([f]\) of functions that agree almost everywhere.
Following universal convention, we write \(f \in L^p\) rather than \([f] \in L^p\), understanding that
"\(f\)" refers to the equivalence class and not to any particular representative. This notational abuse
is harmless because all the quantities we care about are invariant under modification on sets of measure
zero. They include the norm \(\|f\|_p\), integrals \(\int fg \, d\mu\), and convergence statements. When
we write \(f = 0\) in \(L^p\), we mean \(f(x) = 0\) for \(\mu\)-almost every \(x\).
The
\(L^\infty\) space
was introduced alongside the finite-exponent case. We restate it here with the essential
boundedness condition spelled out, since the endpoint \(p = \infty\) occurs in both of the main
inequalities below.
Definition: The \(L^\infty\) Space
Let \((\Omega, \mathcal{F}, \mu)\) be a measure space. A measurable function
\(f : \Omega \to \mathbb{F}\) is essentially bounded if there exists a constant
\(C \geq 0\) such that \(|f(x)| \leq C\) for a.e. \(x\). The essential supremum is
\[
\|f\|_\infty = \inf\{C \geq 0 : |f(x)| \leq C \text{ for a.e. } x\}.
\]
The \(L^\infty\) space is the set of equivalence classes (under a.e.-equality) of
essentially bounded measurable functions, equipped with the quantity \(\|\cdot\|_\infty\). We verify
below that this is indeed a norm.
The infimum in the definition of \(\|f\|_\infty\) is not just an infimum. It is actually attained. This
small fact is what makes \(\|\cdot\|_\infty\) well-behaved as a norm and is used repeatedly in the
sequel.
Lemma: Essential Supremum Is Attained
If \(f \in L^\infty\), then \(|f(x)| \leq \|f\|_\infty\) for almost every \(x\).
In particular, the infimum in the definition of \(\|f\|_\infty\) is achieved.
Proof:
Since \(f \in L^\infty\), the set of admissible bounds is non-empty. It is also bounded below by
\(0\), so its infimum \(\|f\|_\infty\) lies in \([0, \infty)\). By the defining property of the
infimum, for each \(n \in \mathbb{N}\) there exists an admissible \(C_n\) with
\(\|f\|_\infty \leq C_n \lt \|f\|_\infty + 1/n\). For this \(C_n\), the set
\(E_n = \{x : |f(x)| \gt C_n\}\) has measure zero. Let \(E = \bigcup_{n=1}^\infty E_n\). Since each
\(E_n\) is null and \(\mu\) is \(\sigma\)-subadditive (a direct consequence of
countable additivity, via
disjointification), \(\mu(E) \leq \sum_{n=1}^\infty \mu(E_n) = 0\).
For \(x \notin E\), we have \(|f(x)| \leq C_n \lt \|f\|_\infty + 1/n\) for every \(n\). Letting
\(n \to \infty\) gives \(|f(x)| \leq \|f\|_\infty\). This proves the first claim. The second is its
immediate corollary, since \(C = \|f\|_\infty\) is itself an admissible bound.
With this lemma in hand, the three norm axioms for \(\|\cdot\|_\infty\) are immediate, and
non-negativity is clear from the definition. Definiteness on equivalence classes. If
\(\|f\|_\infty = 0\), then \(|f(x)| \leq 0\) a.e., so \(f = 0\) a.e., which says that \([f]\) is the zero
class. Absolute homogeneity. For \(\alpha = 0\) both sides vanish. For \(\alpha \neq 0\), the
lemma gives \(|f(x)| \leq \|f\|_\infty\) a.e., so \(|\alpha f(x)| \leq |\alpha| \|f\|_\infty\) a.e.,
showing \(\|\alpha f\|_\infty \leq |\alpha| \|f\|_\infty\). Applying the same argument to
\(\alpha^{-1}(\alpha f) = f\) yields the reverse inequality. Triangle inequality. Since
\(|f(x)| \leq \|f\|_\infty\) and \(|g(x)| \leq \|g\|_\infty\) a.e., we have
\(|f(x) + g(x)| \leq \|f\|_\infty + \|g\|_\infty\) a.e., which forces
\(\|f + g\|_\infty \leq \|f\|_\infty + \|g\|_\infty\).
So \((L^\infty, \|\cdot\|_\infty)\) is a normed space. Its completeness is established alongside the
\(1 \leq p \lt \infty\) case, by a genuinely simpler argument, when we prove the Riesz-Fischer theorem.
Connection to Sequence Spaces
The sequence spaces \(\ell^p\) are a special case of \(L^p\). If we take \(\Omega = \mathbb{N}\),
\(\mathcal{F} = 2^{\mathbb{N}}\) (all subsets), and \(\mu\) = counting measure (\(\mu(\{n\}) = 1\)
for each \(n\)), then \(L^p(\mathbb{N}, \mu)\) is exactly \(\ell^p\). In this setting, the
equivalence class issue is trivial. Since every singleton \(\{n\}\) has positive measure, two
sequences are equal a.e. if and only if they are identical. Every theorem we prove for \(L^p\) on
this page therefore specializes to \(\ell^p\) automatically.
With the quotient construction in hand, the seminorm \(\|\cdot\|_p\) on \(\mathscr{L}^p\) descends to a
genuine norm on \(L^p\). If \(\|[f]\|_p = 0\), then \(f = 0\) a.e., which means \([f]\) is the
zero element of the quotient space. The remaining norm axiom is the triangle inequality
\(\|f + g\|_p \leq \|f\|_p + \|g\|_p\). It is the content of Minkowski's inequality,
which requires Hölder's inequality as an intermediate step. We turn to these now.
Young's & Hölder's Inequality
The proof chain begins with an elementary inequality about real numbers,
which we then "integrate" to obtain the central inequality of \(L^p\) theory.
Hölder Conjugates
Definition: Hölder Conjugate
For an exponent \(1 \lt p \lt \infty\), the Hölder conjugate \(q\) is the
unique real number satisfying
\[
\begin{align*}
\frac{1}{p} + \frac{1}{q} &= 1, \\\\
\text{equivalently} \quad q &= \frac{p}{p - 1}.
\end{align*}
\]
Since \(p \gt 1\), \(p - 1 \gt 0\) gives \(q \gt 0\). Also \(p \gt p - 1\) gives \(q \gt 1\), while
\(p \lt \infty\) gives \(q \lt \infty\). Thus \(1 \lt q \lt \infty\).
The extreme cases are defined by convention so that the relation \(1/p + 1/q = 1\) remains formally
valid with the extended-real convention \(1/\infty = 0\). The conjugate of \(p = 1\) is
\(q = \infty\), and the conjugate of \(p = \infty\) is \(q = 1\). The relation is symmetric, so the
conjugate of \(q\) is \(p\).
The unique self-conjugate case is \(p = q = 2\), the setting of
Hilbert spaces and
the
Cauchy-Schwarz inequality.
The algebraic identity \((p - 1)q = p\), which follows immediately from \(q = p/(p-1)\), will appear
repeatedly in the proofs below.
Young's Inequality
Theorem: Young's Inequality
Let \(1 \lt p \lt \infty\) and let \(q\) be its Hölder conjugate. For all \(a, b \geq 0\),
\[
ab \leq \frac{a^p}{p} + \frac{b^q}{q}.
\]
Equality holds if and only if \(a^p = b^q\).
Proof:
The argument is a direct specialization of the weighted AM-GM inequality with weights \((1/p, 1/q)\).
If \(a = 0\) or \(b = 0\), both sides reduce to a non-negative quantity and the inequality holds
trivially (see the equality analysis below). Assume \(a, b \gt 0\).
The key observation is that the logarithm \(t \mapsto \log t\) is strictly
concave on \((0, \infty)\). Equivalently, \(-\log\) is
strictly convex
there, which the
second-order condition for convexity
gives at once on the open convex set \((0, \infty) \subseteq \mathbb{R}\), where \(-\log\) is
\(C^2\) with \((-\log t)'' = 1/t^2 \gt 0\). Unwinding this, for any \(\lambda \in (0, 1)\) and
\(u, v \gt 0\) with \(u \neq v\),
\[
\log\bigl(\lambda u + (1 - \lambda) v\bigr) \gt \lambda \log u + (1 - \lambda) \log v,
\]
with equality when \(u = v\).
By the logarithm rules
\(\lambda \log u + (1 - \lambda) \log v = \log\bigl(u^\lambda v^{1-\lambda}\bigr)\). Since \(\log\)
is strictly increasing, we obtain the weighted AM-GM inequality
\[
u^\lambda \, v^{1 - \lambda} \leq \lambda u + (1 - \lambda) v,
\]
with equality if and only if \(u = v\).
Now set \(\lambda = 1/p\) (so that \(1 - \lambda = 1/q\) and \(\lambda \in (0, 1)\) since
\(1 \lt p \lt \infty\)), \(u = a^p\), and \(v = b^q\):
\[
(a^p)^{1/p} (b^q)^{1/q} \leq \frac{a^p}{p} + \frac{b^q}{q}.
\]
Simplifying the left side, \((a^p)^{1/p} (b^q)^{1/q} = a \cdot b\), which gives
\(ab \leq \frac{a^p}{p} + \frac{b^q}{q}\).
Equality condition. Take first \(a, b \gt 0\). The weighted AM-GM step used strict
concavity of \(\log\) with \(\lambda = 1/p \in (0, 1)\), so equality holds if and only if \(u = v\),
that is, \(a^p = b^q\). It remains to check the degenerate cases. If \(a = 0\) and \(b \gt 0\), the
inequality reads \(0 \leq b^q/q\), which is strict (since \(b^q/q \gt 0\)), and correspondingly
\(a^p = 0 \neq b^q\). The case \(b = 0\), \(a \gt 0\) is symmetric. If \(a = b = 0\), both sides are
zero and \(a^p = 0 = b^q\). Hence in every case, equality holds if and only if \(a^p = b^q\).
Young's inequality is a pointwise statement about real numbers. Its power emerges when we "integrate both
sides", which is exactly what happens in the proof of Hölder's inequality.
Hölder's Inequality
Theorem: Hölder's Inequality
Let \(1 \leq p \leq \infty\) and let \(q\) be its Hölder conjugate. If \(f \in L^p\) and
\(g \in L^q\), then the product \(fg\) is measurable, \(fg \in L^1\), and
\[
\begin{align*}
\|fg\|_1
&= \int_\Omega |f g| \, d\mu \\\\
&\leq \|f\|_p \, \|g\|_q.
\end{align*}
\]
Proof:
Measurability of \(fg\) is standard, since the product of measurable functions is measurable (in the
complex case, by considering real and imaginary parts). We establish the inequality below. Because
\(\|f\|_p, \|g\|_q \lt \infty\) by hypothesis, its right-hand side is finite, which gives the
integrability conclusion \(fg \in L^1\).
Case \(p = 1, q = \infty\) (the case \(p = \infty, q = 1\) follows by interchanging the roles
of \(f\) and \(g\)):
If \(\|f\|_1 = 0\), then by
Theorem: Zero Integral Implies Vanishing Almost Everywhere,
\(|f| = 0\) a.e., so \(fg = 0\) a.e. and both sides are zero. If \(\|g\|_\infty = 0\), then by
Lemma: Essential Supremum Is Attained,
\(|g(x)| \leq 0\) a.e., so \(g = 0\) a.e. and again \(fg = 0\) a.e. Otherwise, by the same lemma,
\(|g(x)| \leq \|g\|_\infty\) for a.e. \(x\). Using multiplicativity of the absolute value,
\[
\begin{align*}
|f(x) g(x)|
&= |f(x)| \cdot |g(x)| \\\\
&\leq |f(x)| \cdot \|g\|_\infty \text{ for a.e. } x.
\end{align*}
\]
Integrating both sides over \(\Omega\) (using monotonicity of the integral for a.e. inequality, and
linearity to pull out the constant \(\|g\|_\infty\)),
\[
\begin{align*}
\int_\Omega |fg| \, d\mu
&\leq \int_\Omega |f| \cdot \|g\|_\infty \, d\mu \\\\
&= \|g\|_\infty \int_\Omega |f| \, d\mu \\\\
&= \|f\|_1 \|g\|_\infty.
\end{align*}
\]
Case \(1 \lt p \lt \infty\):
If \(\|f\|_p = 0\) or \(\|g\|_q = 0\), then \(f = 0\) a.e. or \(g = 0\) a.e., so \(fg = 0\) a.e. and
both sides are zero. Assume \(\|f\|_p \gt 0\) and \(\|g\|_q \gt 0\).
Normalization.
The strategy is to rescale \(f\) and \(g\) to have unit norm, reducing the integral inequality to a
pointwise application of Young's inequality. Define
\[
\begin{align*}
\tilde{f} &= \frac{|f|}{\|f\|_p}, \\\\
\tilde{g} &= \frac{|g|}{\|g\|_q}.
\end{align*}
\]
By direct computation,
\[
\begin{align*}
\|\tilde f\|_p^p
&= \int |\tilde f|^p \, d\mu \\\\
&= \int |f|^p / \|f\|_p^p \, d\mu \\\\
&= \|f\|_p^p / \|f\|_p^p \\\\
&= 1,
\end{align*}
\]
so \(\|\tilde f\|_p = 1\), and similarly \(\|\tilde g\|_q = 1\).
Since \(f, g\) take values in \(\mathbb{F}\), we have \(\tilde{f}(x), \tilde{g}(x) \in [0, \infty)\)
everywhere, and
Young's inequality applies
pointwise:
\[
\tilde{f}(x) \, \tilde{g}(x)
\leq \frac{\tilde{f}(x)^p}{p} + \frac{\tilde{g}(x)^q}{q}
\quad \text{for every } x.
\]
Integrating this pointwise inequality over \(\Omega\) (by monotonicity of the integral) and using
linearity,
\[
\begin{align*}
\int_\Omega \tilde{f} \, \tilde{g} \, d\mu
&\leq \frac{1}{p} \int_\Omega \tilde{f}^p \, d\mu + \frac{1}{q} \int_\Omega \tilde{g}^q \, d\mu \\\\
&= \frac{1}{p} + \frac{1}{q} \\\\
&= 1.
\end{align*}
\]
Substituting back \(\tilde{f} = |f|/\|f\|_p\) and \(\tilde{g} = |g|/\|g\|_q\):
\[
\frac{1}{\|f\|_p \, \|g\|_q} \int_\Omega |f g| \, d\mu \leq 1,
\]
which gives \(\int |fg| \, d\mu \leq \|f\|_p \, \|g\|_q\) as claimed.
Equality Conditions and Special Cases
Tracing through the proof reveals when Hölder's inequality is sharp.
Equality Condition for Hölder (\(1 \lt p \lt \infty\))
Let \(1 \lt p \lt \infty\), let \(q\) be its Hölder conjugate, and let
\(f \in L^p\), \(g \in L^q\). Then equality holds in Hölder's inequality,
\[
\int_\Omega |f g| \, d\mu = \|f\|_p \, \|g\|_q,
\]
if and only if either \(f = 0\) a.e., or \(g = 0\) a.e., or there is a constant
\(c \gt 0\) with \(|f|^p = c \, |g|^q\) a.e.
Proof:
Assume first that \(\|f\|_p \gt 0\) and \(\|g\|_q \gt 0\) (the degenerate cases \(f = 0\) or
\(g = 0\) a.e. give equality trivially, with both sides zero). Using the normalized
\(\tilde f = |f|/\|f\|_p\) and \(\tilde g = |g|/\|g\|_q\) from the main proof, define the
pointwise gap
\[
\Delta(x) = \frac{\tilde f(x)^p}{p} + \frac{\tilde g(x)^q}{q}
- \tilde f(x)\tilde g(x).
\]
Since \(\tilde f(x), \tilde g(x) \in [0, \infty)\) for every \(x\), pointwise Young's
inequality gives \(\Delta(x) \geq 0\) for every \(x\), and \(\Delta\) is measurable as a sum
of compositions of measurable functions. It is therefore an admissible integrand for
Theorem: Zero Integral Implies Vanishing Almost Everywhere.
The main proof's integration step rewritten in equation form reads
\[
\begin{align*}
\int \tilde f \, \tilde g \, d\mu
&= \tfrac{1}{p} + \tfrac{1}{q} - \int \Delta \, d\mu \\\\
&= 1 - \int \Delta \, d\mu.
\end{align*}
\]
Equality in Hölder's inequality means \(\int \tilde f \tilde g \, d\mu = 1\), which is
therefore equivalent to \(\int \Delta \, d\mu = 0\). By the zero-integral theorem just
invoked, \(\Delta = 0\) a.e. By the equality case of
Young's inequality,
\(\Delta(x) = 0\) forces \(\tilde f(x)^p = \tilde g(x)^q\) for a.e. \(x\).
Unwinding the normalization,
\[
\frac{|f(x)|^p}{\|f\|_p^p} = \frac{|g(x)|^q}{\|g\|_q^q}
\quad \text{for a.e. } x,
\]
so \(|f|^p\) and \(|g|^q\) are proportional a.e. There exists a constant
\(c = \|f\|_p^p / \|g\|_q^q \gt 0\) with \(|f|^p = c \, |g|^q\) a.e.
Conversely, suppose \(|f|^p = c |g|^q\) a.e. for some \(c \gt 0\). Integrating both sides gives
\(\|f\|_p^p = c \, \|g\|_q^q\), so \(c\) is forced to equal \(\|f\|_p^p / \|g\|_q^q\). Then
\(\tilde f^p = \tilde g^q\) a.e., so \(\Delta = 0\) a.e. and \(\int \Delta \, d\mu = 0\). The main
proof's integration identity then gives
\[
\int \tilde f \, \tilde g \, d\mu = 1 - 0 = 1.
\]
Unwinding the normalization gives \(\int |fg| \, d\mu = \|f\|_p \, \|g\|_q\), that is, equality in
Hölder's inequality.
When \(p = q = 2\), Hölder's inequality reduces to:
\[
\int_\Omega |fg| \, d\mu \leq \|f\|_2 \, \|g\|_2.
\]
For the standard \(L^2\) inner product \(\langle f, g \rangle = \int f \, \overline{g} \, d\mu\), the
triangle inequality for integrals gives \(|\langle f, g \rangle| \leq \int |fg| \, d\mu\), so
Hölder's inequality immediately implies the
Cauchy-Schwarz inequality
\(|\langle f, g \rangle| \leq \|f\|_2 \|g\|_2\). The finite-dimensional bound
\(|\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\| \, \|\mathbf{v}\|\) on
Hilbert spaces is the
special case of counting measure on a finite set. Hölder's inequality thus extends Cauchy-Schwarz to
the full family of conjugate exponent pairs, in which \(p = q = 2\) is the self-conjugate and most
symmetric member.
The One Direction of \(L^p\) Duality
Hölder's inequality immediately settles "one half" of the duality claim from
Dual Spaces. For any fixed
\(g \in L^q\), the map
\[
\begin{align*}
&\varphi_g : L^p \to \mathbb{F}, \\\\
&\varphi_g(f) = \int_\Omega f \, g \, d\mu
\end{align*}
\]
is a
bounded linear functional
on \(L^p\). Two facts this presupposes are still outstanding: that
\(\alpha f_1 + \beta f_2 \in L^p\) for \(f_1, f_2 \in L^p\) and \(\alpha, \beta \in \mathbb{F}\),
and that \(\|\cdot\|_p\) satisfies the triangle inequality, so that \(L^p\) is a normed vector
space. Both are established in the next section via Minkowski's inequality. We state the duality
result here because Hölder's inequality is its main analytic ingredient.
The map \(\varphi_g\) is well-defined on equivalence classes. If \(f \sim f'\) (that is, \(f = f'\)
a.e.), then \(fg = f'g\) a.e., so \(\int fg \, d\mu = \int f'g \, d\mu\). Linearity in \(f\) follows from
linearity of the integral:
\(\varphi_g(\alpha f_1 + \beta f_2) = \alpha \varphi_g(f_1) + \beta \varphi_g(f_2)\). Boundedness is
Hölder: \(\|\varphi_g\|_{(L^p)^*} = \sup_{\|f\|_p = 1} |\varphi_g(f)| \leq \|g\|_q\).
This bound is in fact sharp. We construct, for each case, a
normalized \(f\) witnessing (or approaching) the equality
\(|\varphi_g(f)| = \|g\|_q\).
Case \(1 \lt p \lt \infty\):
For \(g \neq 0\), define the "equalizer"
\[
f_0(x) = \|g\|_q^{\,1-q} \,|g(x)|^{q-1} \, \overline{\operatorname{sgn} g(x)}
\]
where \(\operatorname{sgn} z = z/|z|\) for \(z \neq 0\) (the complex "phase", equal to \(\pm 1\) or \(0\)
in the real case) and \(\operatorname{sgn} 0 = 0\).
Each factor is measurable. The quantity \(\|g\|_q^{1-q}\) is a constant. Next, \(|g|^{q-1}\) is the
composition of the measurable function \(|g|\) with the Borel map \(t \mapsto t^{q-1}\) on
\([0, \infty)\). Finally, \(\operatorname{sgn} g\) is the composition of \(g\) with the Borel map
\(z \mapsto \operatorname{sgn} z\) on \(\mathbb{F}\) (continuous on \(\{z \neq 0\}\), extended by
\(\operatorname{sgn} 0 = 0\)). Hence \(f_0\) is measurable. Using the algebraic identity \(p(q - 1) = q\)
(equivalently, \(p(1 - q) = -q\)),
\[
\begin{align*}
|f_0|^p
&= \|g\|_q^{p(1-q)} \, |g|^{p(q-1)} \\\\
&= \|g\|_q^{-q} \, |g|^q,
\end{align*}
\]
so \(\int |f_0|^p \, d\mu = \|g\|_q^{-q} \int |g|^q \, d\mu = \|g\|_q^{-q} \cdot \|g\|_q^q = 1\). In
particular \(f_0 \in L^p\) with \(\|f_0\|_p = 1\). The identity
\(|f_0|^p = \|g\|_q^{-q} |g|^q\) is exactly the
equality condition for Hölder's inequality
with \(c = \|g\|_q^{-q} \gt 0\), so the pair \((f_0, g)\) is extremal by construction.
This is what dictates the exponent \(q - 1\) in the definition of \(f_0\). Moreover
\(f_0(x) g(x) = \|g\|_q^{1-q} |g(x)|^q \geq 0\), so
\(\varphi_g(f_0) = \|g\|_q^{1-q} \int |g|^q \, d\mu
= \|g\|_q^{1-q} \cdot \|g\|_q^q = \|g\|_q\).
Case \(p = \infty, q = 1\):
For \(g \neq 0\), take \(f_0 = \overline{\operatorname{sgn} g}\). Then \(|f_0| \leq 1\) pointwise with
\(|f_0| = 1\) on \(\{g \neq 0\}\), so \(\|f_0\|_\infty = 1\). Also \(f_0 g = |g|\), so
\(\varphi_g(f_0) = \int |g| \, d\mu = \|g\|_1\).
Case \(p = 1, q = \infty\):
Assuming \(\mu\) is
\(\sigma\)-finite,
the bound \(\|g\|_\infty\) need not be attained by a single \(f\), but it is approached. For
\(g \neq 0\) and any \(\varepsilon\) with \(0 \lt \varepsilon \lt \|g\|_\infty\), the set
\(E_\varepsilon = \{x : |g(x)| \gt \|g\|_\infty - \varepsilon\}\) has positive measure by the
definition of \(\|g\|_\infty\) as an infimum. By \(\sigma\)-finiteness, choose
\(F_\varepsilon \subseteq E_\varepsilon\) with \(0 \lt \mu(F_\varepsilon) \lt \infty\). Set
\(f_\varepsilon = \mu(F_\varepsilon)^{-1} \chi_{F_\varepsilon} \, \overline{\operatorname{sgn} g}\).
Then \(\|f_\varepsilon\|_1 = 1\) and
\[
\begin{align*}
\varphi_g(f_\varepsilon)
&= \mu(F_\varepsilon)^{-1} \int_{F_\varepsilon} |g| \, d\mu \\\\
&\geq \|g\|_\infty - \varepsilon.
\end{align*}
\]
Letting \(\varepsilon \to 0^+\), \(\sup_{\|f\|_1 = 1} |\varphi_g(f)| \geq \|g\|_\infty\). (Without
\(\sigma\)-finiteness, this bound can fail to be attained, and the full duality
\((L^1)^* \cong L^\infty\) breaks down. The exact failure lies beyond the scope of this page.)
Combining the three cases with the upper bound from Hölder's inequality (the case \(g = 0\)
being trivial, since both sides vanish), \(\|\varphi_g\|_{(L^p)^*} = \|g\|_q\) for all
\(1 \leq p \leq \infty\) (with \(\sigma\)-finiteness for \(p = 1\)), so the embedding
\(L^q \hookrightarrow (L^p)^*\) given by \(g \mapsto \varphi_g\) is isometric.
Every element of \(L^q\) gives rise to a continuous functional on \(L^p\), with no loss of norm.
The converse is substantially harder and relies on the
Radon-Nikodym theorem.
This direction asserts that every continuous functional on \(L^p\) arises from some
\(g \in L^q\), for \(1 \leq p \lt \infty\), with \(\sigma\)-finiteness assumed when \(p = 1\). The
standard construction, carried out for \(\sigma\)-finite \(\mu\) and real scalars, evaluates a
continuous functional \(\Lambda\) on \(L^p(\mu)\) at the indicators \(\chi_A\) of sets of finite
measure, assembles the resulting values into a
signed measure
\(\nu\) with \(\nu \ll \mu\), passes to the two non-negative measures of its
Jordan decomposition,
and extracts \(g = d\nu/d\mu\) as the representing element. The complex case follows by treating
real and imaginary parts separately.
Minkowski's Inequality & the \(L^p\) Norm
We now use Hölder's inequality to prove the triangle inequality for \(\|\cdot\|_p\), completing the
verification that \(L^p\) is a normed space.
Theorem: Minkowski's Inequality
Let \(1 \leq p \leq \infty\). If \(f, g \in L^p\), then \(f + g \in L^p\) and
\[
\|f + g\|_p \leq \|f\|_p + \|g\|_p.
\]
Proof:
Throughout, \(f + g\) is measurable as the sum of measurable functions (componentwise in the complex
case), so \(|f + g|\) and its powers are measurable as well. The proof splits into three cases. The
endpoints \(p = 1\) and \(p = \infty\) follow from the pointwise triangle inequality and standard
properties of integration. The main case \(1 \lt p \lt \infty\) carries the substance of the argument
and uses Hölder's inequality.
Case \(p = 1\):
By the pointwise triangle inequality, \(|f(x) + g(x)| \leq |f(x)| + |g(x)|\). Integrating this a.e.
inequality over \(\Omega\) (using monotonicity of the integral, with \(f + g\) measurable as
established above),
\[
\begin{align*}
\int_\Omega |f + g| \, d\mu
&\leq \int_\Omega |f| \, d\mu + \int_\Omega |g| \, d\mu \\\\
&= \|f\|_1 + \|g\|_1 \\\\
&\lt \infty.
\end{align*}
\]
Hence \(f + g \in L^1\), and by the definition of the \(L^1\) norm,
\(\|f+g\|_1 = \int |f+g| \, d\mu \leq \|f\|_1 + \|g\|_1\).
Case \(p = \infty\):
By
Lemma: Essential Supremum Is Attained,
\(|f(x)| \leq \|f\|_\infty\) and \(|g(x)| \leq \|g\|_\infty\) for a.e. \(x\), whence
\(|f(x) + g(x)| \leq \|f\|_\infty + \|g\|_\infty \lt \infty\) for a.e. \(x\). This shows \(f + g\) is
essentially bounded, that is, \(f + g \in L^\infty\). The constant \(\|f\|_\infty + \|g\|_\infty\) is
an admissible a.e.-bound for \(|f + g|\), so by the definition of \(\|f + g\|_\infty\) as the infimum
of such bounds, \(\|f + g\|_\infty \leq \|f\|_\infty + \|g\|_\infty\).
Case \(1 \lt p \lt \infty\):
First, observe that \(f + g \in L^p\). Pointwise,
\(|f(x) + g(x)| \leq |f(x)| + |g(x)| \leq 2\max(|f(x)|, |g(x)|)\), and since \(t \mapsto t^p\) is
monotonically increasing on \([0, \infty)\) for \(p \gt 0\),
\[
\begin{align*}
|f + g|^p
&\leq \bigl(2 \max(|f|, |g|)\bigr)^p \\\\
&= 2^p \max(|f|, |g|)^p \\\\
&= 2^p \max(|f|^p, |g|^p) \\\\
&\leq 2^p (|f|^p + |g|^p).
\end{align*}
\]
Integrating, \(\int |f+g|^p \, d\mu \leq 2^p(\|f\|_p^p + \|g\|_p^p) \lt \infty\), so
\(f + g \in L^p\) and \(\|f+g\|_p = \bigl(\int |f+g|^p \, d\mu\bigr)^{1/p}\) is finite.
Now assume \(\|f + g\|_p \gt 0\) (otherwise the inequality is trivial). We begin by splitting
\(|f + g|^p\). Since \(|f+g|^{p-1} \geq 0\), multiplying the pointwise triangle inequality
\(|f+g| \leq |f| + |g|\) by \(|f+g|^{p-1}\) preserves its direction, giving
\[
\begin{align*}
|f + g|^p
&= |f + g|^{p-1} \cdot |f + g| \\\\
&\leq |f + g|^{p-1} |f| + |f + g|^{p-1} |g|.
\end{align*}
\]
We now apply Hölder's inequality to
each term on the right. Each of \(|f+g|^{p-1}\), \(|f|\), \(|g|\) is non-negative and measurable,
since \(|f+g|^{p-1}\) arises by composing \(|f+g|\) with the Borel map \(t \mapsto t^{p-1}\) on
\([0, \infty)\), and similarly for \(|f|, |g|\).
It remains to verify the relevant \(L^p\) memberships. We have \(|f| \in L^p\) directly (since
\(\int |f|^p \, d\mu = \|f\|_p^p \lt \infty\)), and similarly \(|g| \in L^p\). For
\(|f+g|^{p-1} \in L^q\),
\[
\begin{align*}
\int_\Omega \bigl(|f + g|^{p-1}\bigr)^q \, d\mu
&= \int_\Omega |f + g|^{(p-1)q} \, d\mu \\\\
&= \int_\Omega |f + g|^{p} \, d\mu \\\\
&= \|f + g\|_p^p,
\end{align*}
\]
where we used the identity \((p - 1)q = p\). Therefore
\(\bigl\| |f+g|^{p-1} \bigr\|_q = \|f+g\|_p^{p/q}\).
Applying Hölder's inequality with non-negative factors (so
\(|(\cdot)(\cdot)| = (\cdot)(\cdot)\)):
\[
\begin{align*}
\int |f + g|^{p-1} |f| \, d\mu
&\leq \bigl\||f+g|^{p-1}\bigr\|_q \cdot \|f\|_p \\\\
&= \|f+g\|_p^{p/q} \cdot \|f\|_p,
\end{align*}
\]
Analogously \(\int |f+g|^{p-1} |g| \, d\mu \leq \|f+g\|_p^{p/q} \cdot \|g\|_p\). Adding these two
bounds and using linearity of the integral to combine the left-hand sides,
\[
\begin{align*}
\|f + g\|_p^p
= \int |f+g|^p \, d\mu
&\leq \int |f+g|^{p-1} |f| \, d\mu + \int |f+g|^{p-1} |g| \, d\mu \\\\
&\leq \|f+g\|_p^{p/q} \bigl(\|f\|_p + \|g\|_p\bigr).
\end{align*}
\]
Since \(\|f+g\|_p^{p/q}\) is strictly positive (by assumption) and finite (established above), we may
divide both sides by it:
\[
\|f + g\|_p^{p - p/q} \leq \|f\|_p + \|g\|_p.
\]
Since \(p - p/q = p(1 - 1/q) = p \cdot (1/p) = 1\), the left side is simply \(\|f + g\|_p\),
completing the proof.
Equality Condition for Minkowski (\(1 \lt p \lt \infty\))
Equality \(\|f + g\|_p = \|f\|_p + \|g\|_p\) holds if and only if \(f\) and \(g\) are
non-negatively proportional a.e., that is, either \(f = 0\) a.e., or \(g = 0\) a.e.,
or there exists a constant \(c \gt 0\) such that \(f = c \, g\) a.e. (equivalently, \(g = c^{-1} f\)
a.e.).
Proof:
Excluding the trivial cases where \(f = 0\) or \(g = 0\) a.e., assume \(\|f\|_p, \|g\|_p \gt 0\).
Then by the just-proved Minkowski inequality, \(\|f + g\|_p \leq \|f\|_p + \|g\|_p\). Moreover, under
our equality hypothesis \(\|f + g\|_p = \|f\|_p + \|g\|_p\), we have \(\|f + g\|_p \gt 0\) (since
\(\|f\|_p, \|g\|_p \gt 0\) by assumption). Tracing the main proof, equality in Minkowski requires
equality in both ingredients used:
- The pointwise triangle inequality \(|f + g| \leq |f| + |g|\) must hold with equality a.e.
on \(\{|f+g| \gt 0\}\), that is, \(|f(x) + g(x)| = |f(x)| + |g(x)|\) for a.e. such \(x\).
(The main proof multiplies this inequality by \(|f+g|^{p-1}\), which vanishes on
\(\{|f+g| = 0\}\), so nothing follows there. Item (2) closes that set.) Equality in this
triangle inequality occurs if and only if \(f(x), g(x)\) lie on a common non-negative ray
from the origin. Explicitly, for \(\mathbb{F} = \mathbb{R}\), either one of \(f(x), g(x)\)
is zero or they share the same sign. For \(\mathbb{F} = \mathbb{C}\), either one is zero
or they have the same complex argument (phase).
- Each of the two Hölder applications must hold with equality. By the
equality condition for Hölder's inequality,
applied with exponents \((p, q)\) to the factors \(|f+g|^{p-1}\) (in \(L^q\))
and \(|f|\) (in \(L^p\)), there exists \(\alpha \gt 0\) with
\[
\begin{align*}
|f|^p
&= \alpha \cdot \bigl(|f+g|^{p-1}\bigr)^q \\\\
&= \alpha |f+g|^{(p-1)q} \\\\
&= \alpha |f+g|^p \quad\text{a.e.},
\end{align*}
\]
using \((p-1)q = p\). Similarly, from the second Hölder application there exists \(\beta \gt 0\) with
\[
|g|^p = \beta |f+g|^p \quad\text{a.e.}
\]
On the set \(\{|f+g| = 0\}\), both relations give \(|f|^p = 0 = |g|^p\) a.e.,
so \(f = g = 0\) a.e. on this set. The desired proportionality \(|f|^p = (\alpha/\beta) |g|^p\)
then holds trivially there. On \(\{|f+g| \gt 0\}\), we have \(|g|^p = \beta |f+g|^p \gt 0\), so the
ratio \(|f|^p / |g|^p = \alpha/\beta\) is well-defined.
Setting \(k = \alpha/\beta \gt 0\), we conclude
\[
\begin{align*}
|f|^p &= k \, |g|^p \quad\text{a.e.}, \\\\
\text{that is,} \quad |f| &= k^{1/p} |g| \quad\text{a.e.}
\end{align*}
\]
Item (1) gives a common phase a.e. on \(\{|f+g| \gt 0\}\), and item (2) gives
\(|f| = c \, |g|\) a.e. with \(c = k^{1/p} \gt 0\). Combining the two, we obtain
\(f = c \, g\) a.e. Indeed, work modulo a null set. At points where \(g(x) = 0\), (2) gives
\(|f(x)| = 0\), so \(f(x) = 0 = c \cdot g(x)\). At points where \(g(x) \neq 0\), the relation
\(|g|^p = \beta |f+g|^p\) from (2) places \(x\) in \(\{|f+g| \gt 0\}\), where (1) applies and
forces \(f(x)\) to have the same phase as \(g(x)\). Then (2) fixes \(|f(x)| = c \, |g(x)|\),
so \(f(x) = c \, g(x)\).
Conversely, if \(f = c \, g\) a.e. with \(c \gt 0\), then \(f + g = (c + 1) g\) a.e., so by absolute
homogeneity \(\|f + g\|_p = (c + 1) \|g\|_p\), and
\(\|f\|_p + \|g\|_p = c \|g\|_p + \|g\|_p = (c + 1) \|g\|_p = \|f + g\|_p\).
Geometrically, Minkowski's inequality is strict whenever \(f\) and \(g\) point in genuinely different
"directions" in \(L^p\). This is a manifestation of the strict convexity of the \(L^p\)
norm for \(1 \lt p \lt \infty\). (At the endpoints \(p = 1\) and \(p = \infty\), strict convexity fails
and equality can occur in many more configurations.)
\(L^p\) Is a Normed Space
Before listing the norm axioms, we record that \(L^p\) is a vector space over
\(\mathbb{F}\). For \(f, g \in L^p\) and \(\alpha, \beta \in \mathbb{F}\), we have
\(\alpha f + \beta g \in L^p\). Closure under addition follows from Minkowski's inequality, case by case
(\(p = 1\), \(p = \infty\), and \(1 \lt p \lt \infty\) were each established as the first step of the
corresponding case in the proof above). Closure under scalar multiplication is immediate from
\(\int |\alpha f|^p \, d\mu = |\alpha|^p \int |f|^p \, d\mu \lt \infty\) (and analogously
\(\|\alpha f\|_\infty = |\alpha| \|f\|_\infty \lt \infty\)).
The quotient construction of \(L^p\) as \(\mathscr{L}^p / \sim\) preserves these operations, since
modifications on null sets are compatible with pointwise addition and scalar multiplication.
We now summarize the complete verification of the norm axioms for \(\|\cdot\|_p\) on
\(L^p(\Omega, \mathcal{F}, \mu)\) in the case \(1 \leq p \lt \infty\). The parallel verification for the
\(p = \infty\) case was completed immediately after
Lemma: Essential Supremum Is Attained.
- Positive definiteness. We have \(\|f\|_p \geq 0\), and
\(\|f\|_p = 0 \iff f = 0\) in \(L^p\) (that is, \(f = 0\) a.e.).
This is where the equivalence class construction is essential.
Without it, \(\|\cdot\|_p\) would only be a seminorm.
- Absolute homogeneity. We have \(\|\alpha f\|_p = |\alpha| \cdot \|f\|_p\) for all
\(\alpha \in \mathbb{F}\). This follows immediately from
\(\int |\alpha f|^p = |\alpha|^p \int |f|^p\).
- Triangle inequality. We have \(\|f + g\|_p \leq \|f\|_p + \|g\|_p\).
This is Minkowski's inequality, proven above.
Therefore \(\bigl(L^p(\Omega, \mathcal{F}, \mu),\, \|\cdot\|_p\bigr)\) is a
normed vector space
for every \(1 \leq p \leq \infty\). This completes our construction of \(L^p\) as a normed structure.
The deepest question, however, remains open: is this normed space
complete? Does every Cauchy sequence in \(L^p\)
converge to a limit that is itself in \(L^p\)? An affirmative answer, the Riesz-Fischer
theorem, elevates \(L^p\) from a normed space to a
Banach space, and
provides the convergence theory that makes \(L^p\) the natural setting for probability, signal
processing, and quantum mechanics. We develop this on
\(L^p\) Completeness & Convergence.