Partitions of Unity

On the open set \(\{t > 0\}\) the function is a composition of smooth functions and hence smooth there; on \(\{t < 0\}\) it is identically zero, hence smooth there. The only point in question is the origin, where we must show that all left- and right-hand derivatives exist and agree at the value \(0\). Two facts about the growth of \(e^{-1/t}\) drive the argument.

First, a decay estimate. For every integer \(k \ge 0\), \[ \lim_{t \searrow 0} \frac{e^{-1/t}}{t^k} = 0. \] Substituting \(s = 1/t\), so that \(s \to +\infty\) as \(t \searrow 0\), this becomes \(\lim_{s \to \infty} s^k e^{-s} = \lim_{s \to \infty} s^k / e^{s}\), which is zero because the exponential dominates every power — a fact established by \(k\) applications of L'Hôpital's rule, or directly from the series \(e^{s} > s^{k+1}/(k+1)!\). Thus \(e^{-1/t}\) tends to zero faster than any power of \(t\) as \(t \searrow 0\).

Second, the form of the derivatives on \(\{t > 0\}\). We claim that for each \(k \ge 0\) there is a polynomial \(p_k\) of degree at most \(k\) such that, for \(t > 0\), \[ f^{(k)}(t) = \frac{p_k(t)\, e^{-1/t}}{t^{2k}}. \] This holds for \(k = 0\) with \(p_0 \equiv 1\). Assuming it for \(k\), differentiate using \(\frac{d}{dt} e^{-1/t} = t^{-2} e^{-1/t}\) and the quotient rule: \[ f^{(k+1)}(t) = \frac{\bigl[t^2 p_k'(t) - 2k\, t\, p_k(t) + p_k(t)\bigr] e^{-1/t}}{t^{2k+2}}, \] so the bracketed expression is a polynomial of degree at most \(k + 1\), establishing the claim for \(k + 1\) by induction. (The precise coefficients are immaterial; only the polynomial-over-power form matters.)

Combining the two. We show by induction that \(f^{(k)}(0) = 0\) for all \(k\), with each derivative continuous at \(0\). For \(k = 0\), continuity at the origin is the statement \(\lim_{t \searrow 0} e^{-1/t} = 0\), which is the decay estimate with \(k = 0\). Suppose \(f^{(k)}(0) = 0\). The right-hand derivative of \(f^{(k)}\) at \(0\) is \[ \lim_{t \searrow 0} \frac{f^{(k)}(t) - f^{(k)}(0)}{t} = \lim_{t \searrow 0} \frac{1}{t}\cdot \frac{p_k(t)\, e^{-1/t}}{t^{2k}} = \lim_{t \searrow 0} p_k(t)\,\frac{e^{-1/t}}{t^{2k+1}}. \] Writing \(p_k(t) = \sum_{j} c_j t^j\), each term is \(c_j\, e^{-1/t} / t^{\,2k + 1 - j}\), and since \(j \le k\) the exponent \(2k + 1 - j\) is a positive integer; by the decay estimate every such term tends to \(0\). The left-hand derivative is \(0\) because \(f\) vanishes identically for \(t \le 0\). The two agree, so \(f^{(k+1)}(0) = 0\). Continuity of \(f^{(k+1)}\) at \(0\) follows from the same estimate applied to its expression on \(\{t > 0\}\). By induction, \(f\) has derivatives of all orders at every point and they are continuous, so \(f\) is smooth, and \(f^{(k)}(0) = 0\) for all \(k\).

Let \(f\) be the seed function of the previous lemma, and define \[ h(t) = \frac{f(r_2 - t)}{f(r_2 - t) + f(t - r_1)}. \] The denominator never vanishes. Indeed, for any \(t\) at least one of the two arguments \(r_2 - t\) and \(t - r_1\) is positive: if both were nonpositive we would have \(t \ge r_2\) and \(t \le r_1\) simultaneously, forcing \(r_2 \le r_1\), contrary to \(r_1 < r_2\). Since \(f\) is nonnegative and strictly positive on positive arguments, the sum \(f(r_2 - t) + f(t - r_1)\) is strictly positive everywhere; \(h\) is therefore a quotient of smooth functions with nonvanishing denominator, hence smooth.

For \(t \ge r_2\) we have \(r_2 - t \le 0\), so \(f(r_2 - t) = 0\) and the numerator vanishes: \(h(t) = 0\). For \(t \le r_1\) we have \(t - r_1 \le 0\), so \(f(t - r_1) = 0\) and the denominator reduces to the numerator: \(h(t) = 1\). For \(r_1 < t < r_2\) both arguments are positive, so both \(f(r_2 - t)\) and \(f(t - r_1)\) are strictly positive, placing the numerator strictly between \(0\) and the denominator; thus \(0 < h(t) < 1\). All the asserted properties hold.

Set \(H(x) = h(|x|)\), where \(h\) is the cutoff function of the previous lemma and \(|x|\) is the Euclidean norm. On \(\mathbb{R}^n \setminus \{0\}\) the norm \(x \mapsto |x|\) is smooth, so \(H\) is there a composition of smooth functions and hence smooth. At the origin a separate argument is needed because the norm is not differentiable there; but \(|x| \le r_1\) precisely on \(\overline{B_{r_1}(0)}\), and on that ball \(h(|x|) \equiv 1\) since \(h \equiv 1\) on \((-\infty, r_1]\). Since \(r_1 > 0\), the open ball \(B_{r_1}(0)\) is a neighborhood of the origin contained in \(\overline{B_{r_1}(0)}\), so \(H\) is identically \(1\) on a neighborhood of the origin, and it is smooth there as well, and therefore smooth on all of \(\mathbb{R}^n\).

The pointwise values transfer directly from \(h\): for \(|x| \le r_1\), \(H(x) = h(|x|) = 1\); for \(|x| \ge r_2\), \(H(x) = h(|x|) = 0\); and for \(r_1 < |x| < r_2\), \(0 < h(|x|) < 1\). In terms of balls these are exactly the three asserted regions, completing the proof.

We give the proof for a smooth manifold without boundary, which is the case we shall use throughout; the construction adapts to manifolds with boundary with no new ideas. The proof proceeds in five movements: refine the cover, build a bump on each piece, sum them, normalize, and reindex.

Step 1 — A refinement by regular coordinate balls. Each \(X_\alpha\), being open in \(M\), is itself a smooth manifold, so by the basis of regular coordinate balls it has a basis \(\mathcal{B}_\alpha\) consisting of regular coordinate balls. The union \(\mathcal{B} = \bigcup_{\alpha} \mathcal{B}_\alpha\) is then a basis for the topology of \(M\): any open subset \(W \subseteq M\) and any point \(q \in W\) lie in some \(X_\alpha\) (the \(X_\alpha\) cover \(M\)), and \(\mathcal{B}_\alpha\), being a basis for \(X_\alpha\), supplies a regular coordinate ball with \(q \in B \subseteq W\). Because \(M\) is a manifold and hence paracompact, the cover \(\mathcal{X}\) admits a countable, locally finite refinement \(\{B_i\}\) all of whose members are drawn from \(\mathcal{B}\) — that is, each \(B_i\) is a regular coordinate ball contained in some \(X_\alpha\). By the closure lemma for locally finite collections, the family of closures \(\{\overline{B_i}\}\) is locally finite as well.

Step 2 — A bump on each ball. Fix \(i\), and let \(X_\alpha\) be a member of the cover containing \(B_i\). Since \(B_i\) is a regular coordinate ball in \(X_\alpha\), its defining property furnishes a slightly larger coordinate ball \(B_i'\) with \(\overline{B_i} \subseteq B_i' \subseteq X_\alpha\) and a smooth coordinate map \(\varphi_i : B_i' \to \mathbb{R}^n\) carrying these onto concentric Euclidean balls, \[ \varphi_i(\overline{B_i}) = \overline{B_{r_i}(0)}, \qquad \varphi_i(B_i') = B_{r_i'}(0), \qquad r_i < r_i'. \] Using the smooth bump function construction with inner and outer radii both taken inside \(B_{r_i}(0)\) — say \(r_i/2\) and \(r_i\) — let \(H_i : \mathbb{R}^n \to \mathbb{R}\) be the resulting smooth function: equal to \(1\) on \(\overline{B_{r_i/2}(0)}\), strictly between \(0\) and \(1\) on the annulus, and \(0\) outside \(B_{r_i}(0)\). In particular \(H_i\) is positive on the open ball \(B_{r_i}(0)\) and zero everywhere outside it, which is all we shall use. Define \(f_i : M \to \mathbb{R}\) by \[ f_i = \begin{cases} H_i \circ \varphi_i & \text{on } B_i',\\[2pt] 0 & \text{on } M \setminus \overline{B_i}. \end{cases} \] These two prescriptions agree wherever their domains overlap, namely on \(B_i' \setminus \overline{B_i}\). Since \(\varphi_i\) is a bijection from \(B_i'\) onto \(B_{r_i'}(0)\) carrying \(\overline{B_i}\) onto \(\overline{B_{r_i}(0)}\), it maps the overlap onto \(B_{r_i'}(0) \setminus \overline{B_{r_i}(0)}\), which lies outside \(B_{r_i}(0)\); and \(H_i\) vanishes outside \(B_{r_i}(0)\). Thus \(H_i \circ \varphi_i = 0\) on the overlap, matching the second prescription, which is zero by fiat. The function \(f_i\) is therefore well-defined, and being locally given by a smooth formula it is smooth on all of \(M\). Its nonzero set is the preimage \(\varphi_i^{-1}(B_{r_i}(0)) = B_i\), so \(\operatorname{supp} f_i = \overline{B_i}\).

Step 3 — Summation. Define \(f : M \to \mathbb{R}\) by \(f = \sum_i f_i\). Because the cover \(\{B_i\}\) is locally finite and \(\operatorname{supp} f_i = \overline{B_i}\), every point of \(M\) has a neighborhood on which all but finitely many \(f_i\) vanish; the sum is therefore locally finite, and \(f\) is smooth. Moreover each \(f_i \ge 0\), and since \(\{B_i\}\) covers \(M\), every point lies in some \(B_i\), where \(f_i > 0\). Hence \(f(p) > 0\) for every \(p \in M\).

Step 4 — Normalization. Define \(g_i = f_i / f\). Since \(f\) is smooth and strictly positive, each \(g_i\) is smooth; clearly \(0 \le g_i \le 1\), and \(\sum_i g_i = (\sum_i f_i)/f = f/f \equiv 1\). The family \((g_i)\) is thus a smooth partition of unity, but indexed by \(i\) rather than by the index set \(A\) of the cover, and subordinate to \(\{B_i\}\) rather than to \(\mathcal{X}\). One adjustment remains.

Step 5 — Reindexing onto \(A\). Because \(\{B_i'\}\) refines \(\mathcal{X}\) — each \(B_i'\) was chosen inside some \(X_\alpha\) in Step 2 — for each \(i\) we may fix an index \(a(i) \in A\) with \(B_i' \subseteq X_{a(i)}\). For each \(\alpha \in A\) define \[ \psi_\alpha = \sum_{i \,:\, a(i) = \alpha} g_i, \] interpreting the sum as the zero function if no \(i\) maps to \(\alpha\). Each \(\psi_\alpha\) is smooth, being a locally finite sum of the smooth \(g_i\), and inherits \(0 \le \psi_\alpha \le 1\). Summing over \(\alpha\) regroups the \(g_i\) without omission, so \(\sum_\alpha \psi_\alpha = \sum_i g_i \equiv 1\), giving condition (iv). For the support condition, the closure lemma permits the closure of a locally finite union to be computed termwise: \[ \operatorname{supp} \psi_\alpha = \overline{\bigcup_{i \,:\, a(i) = \alpha} B_i} = \bigcup_{i \,:\, a(i) = \alpha} \overline{B_i} \subseteq X_\alpha, \] the final inclusion holding because, for each contributing \(i\), \(\overline{B_i} \subseteq B_i' \subseteq X_{a(i)} = X_\alpha\) by the choice of \(B_i'\) in Step 2 and of \(a(i)\) above. This is condition (ii). Finally, the family \((\operatorname{supp} \psi_\alpha)\) is locally finite, since each \(\operatorname{supp} \psi_\alpha\) is a union of members of the locally finite family \(\{\overline{B_i}\}\), giving condition (iii). With (i) already noted, all four conditions hold, and \((\psi_\alpha)_{\alpha \in A}\) is the desired smooth partition of unity subordinate to \(\mathcal{X}\).

The two sets \(U_0 = U\) and \(U_1 = M \setminus A\) form an open cover of \(M\): every point either lies in \(A \subseteq U_0\) or fails to, in which case it lies in \(U_1\). By the existence of partitions of unity, let \(\{\psi_0, \psi_1\}\) be a smooth partition of unity subordinate to this cover, so that \(\operatorname{supp} \psi_0 \subseteq U_0 = U\), \(\operatorname{supp} \psi_1 \subseteq U_1 = M \setminus A\), and \(\psi_0 + \psi_1 \equiv 1\). Since \(\operatorname{supp} \psi_1 \subseteq M \setminus A\), the function \(\psi_1\) vanishes on \(A\); hence on \(A\) the relation \(\psi_0 + \psi_1 = 1\) forces \(\psi_0 \equiv 1\). Together with \(0 \le \psi_0 \le 1\) and \(\operatorname{supp} \psi_0 \subseteq U\), this shows \(\psi_0\) is a smooth bump function for \(A\) supported in \(U\).

Since \(f\) is smooth on \(A\), each point \(p \in A\) has a neighborhood \(W_p\) in \(M\) and a smooth function \(\widetilde{f}_p : W_p \to \mathbb{R}^k\) agreeing with \(f\) on \(W_p \cap A\). Shrinking \(W_p\) to \(W_p \cap U\) if necessary, we may assume \(W_p \subseteq U\). The collection \(\{W_p : p \in A\} \cup \{M \setminus A\}\) is then an open cover of \(M\). Let \(\{\psi_p : p \in A\} \cup \{\psi_0\}\) be a smooth partition of unity subordinate to it, indexed so that \(\operatorname{supp} \psi_p \subseteq W_p\) for each \(p \in A\) and \(\operatorname{supp} \psi_0 \subseteq M \setminus A\).

For each \(p \in A\), the product \(\psi_p \widetilde{f}_p\) is smooth on \(W_p\) and extends smoothly to all of \(M\) when interpreted as \(0\) on \(M \setminus \operatorname{supp} \psi_p\): the two prescriptions agree on the open overlap \(W_p \setminus \operatorname{supp} \psi_p\), where both are zero. Define \[ \widetilde{f} = \sum_{p \in A} \psi_p \widetilde{f}_p. \] Because the supports \(\{\operatorname{supp} \psi_p\}\) are locally finite, this sum has only finitely many nonzero terms near any point and so defines a smooth function on \(M\).

It remains to check that \(\widetilde{f}\) extends \(f\). Fix \(x \in A\). Then \(\psi_0(x) = 0\), since \(\operatorname{supp} \psi_0 \subseteq M \setminus A\); and for each \(p\) with \(\psi_p(x) \ne 0\), the point \(x\) lies in \(\operatorname{supp} \psi_p \subseteq W_p\), so \(\widetilde{f}_p(x) = f(x)\) by the agreement on \(W_p \cap A\). Therefore \[ \widetilde{f}(x) = \sum_{p \in A} \psi_p(x)\,\widetilde{f}_p(x) = \Bigl( \psi_0(x) + \sum_{p \in A} \psi_p(x) \Bigr) f(x) = f(x), \] using \(\psi_0(x) + \sum_p \psi_p(x) = 1\). Hence \(\widetilde{f}|_A = f\). Finally, by the closure lemma, \[ \operatorname{supp} \widetilde{f} \subseteq \overline{\bigcup_{p \in A} \operatorname{supp} \psi_p} = \bigcup_{p \in A} \operatorname{supp} \psi_p \subseteq \bigcup_{p \in A} W_p \subseteq U, \] completing the proof.

A manifold has a countable open cover \(\{V_j\}_{j=1}^\infty\) by precompact sets — sets whose closures \(\overline{V_j}\) are compact — furnished by the basis of precompact coordinate balls, which provides a countable basis of open sets each having compact closure. Let \(\{\psi_j\}\) be a smooth partition of unity subordinate to \(\{V_j\}\), so that \(\operatorname{supp} \psi_j \subseteq V_j\) and \(\sum_j \psi_j \equiv 1\), and define \[ f = \sum_{j=1}^\infty j\, \psi_j. \] The sum is locally finite, so \(f\) is smooth. Since each \(\psi_j \ge 0\) and \(\sum_j \psi_j = 1\), we have \(f = \sum_j j \psi_j \ge \sum_j \psi_j = 1\), so \(f \ge 1\) everywhere; in particular \(f\) is real-valued and continuous.

To see that \(f\) is an exhaustion function, fix \(c \in \mathbb{R}\) and choose any integer \(N \ge c\). We claim that every point \(p\) with \(f(p) \le c\) lies in \(\operatorname{supp} \psi_j\) for some \(j \le N\). Suppose not: then every index \(j\) with \(\psi_j(p) \ne 0\) satisfies \(j > N\), and since the indices are integers this means \(j \ge N + 1\). The indices \(j \le N\) then contribute nothing to \(\sum_j \psi_j(p) = 1\), so \(\sum_{j > N} \psi_j(p) = 1\), and therefore \[ f(p) = \sum_j j\, \psi_j(p) = \sum_{j > N} j\, \psi_j(p) \ge (N + 1) \sum_{j > N} \psi_j(p) = N + 1 > N \ge c, \] contradicting \(f(p) \le c\). Hence \(p\) lies in \(\operatorname{supp} \psi_j \subseteq \overline{V_j}\) for some \(j \le N\), giving \[ f^{-1}\bigl((-\infty, c]\bigr) \subseteq \bigcup_{j \le N} \overline{V_j}, \] a finite union of compact sets, hence compact. The sublevel set is closed, because \(f\) is continuous, and a closed subset of a compact set is compact; so \(f^{-1}((-\infty, c])\) is compact. Since \(c\) was arbitrary, \(f\) is a smooth exhaustion function.