Partitions of Unity

On the open set \(\{t \gt 0\}\) the function is a composition of smooth functions and hence smooth there. On \(\{t \lt 0\}\) it is identically zero, hence smooth as well. 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 \(t \searrow 0\), we rewrite the limit as \(\lim_{s \to \infty} s^k e^{-s} = \lim_{s \to \infty} s^k / e^{s}\), which is zero because the exponential dominates every power. That domination follows from \(k\) applications of L'Hôpital's rule, or directly from the series \(e^{s} \gt s^{k+1}/(k+1)!\). Thus \(e^{-1/t}\) tends to zero faster than any power of \(t\) as \(t \searrow 0\).

Second, the form of the derivatives on \(\{t \gt 0\}\).
We claim that for each \(k \ge 0\) there is a polynomial \(p_k\) of degree at most \(k\) such that, for \(t \gt 0\), \[ f^{(k)}(t) = \frac{p_k(t)\, e^{-1/t}}{t^{2k}}. \] 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 \[ \begin{align*} \lim_{t \searrow 0} \frac{f^{(k)}(t) - f^{(k)}(0)}{t} &= \lim_{t \searrow 0} \frac{1}{t}\cdot \frac{p_k(t)\, e^{-1/t}}{t^{2k}} \\\\ &= \lim_{t \searrow 0} p_k(t)\,\frac{e^{-1/t}}{t^{2k+1}}. \end{align*} \] 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 \(\{t \gt 0\}\). 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 \(r_1 \lt r_2\). Since \(f\) is nonnegative and strictly positive on positive arguments, the sum \(f(r_2 - t) + f(t - r_1)\) is strictly positive everywhere. Therefore \(h\) is 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 \(r_1 \lt t \lt r_2\) 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 \(0 \lt h(t) \lt 1\). 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. However, \(|x| \le r_1\) precisely on \(\overline{B_{r_1}(0)}\), and on that ball \(h(|x|) \equiv 1\) since \(h \equiv 1\) on \((-\infty, r_1]\). Since \(r_1 \gt 0\), 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 hence smooth there as well. Therefore \(H\) is smooth on all of \(\mathbb{R}^n\).

The pointwise values transfer directly from \(h\). For \(|x| \le r_1\) we have \(H(x) = h(|x|) = 1\). For \(|x| \ge r_2\) we have \(H(x) = h(|x|) = 0\). For \(r_1 \lt |x| \lt r_2\) we have \(0 \lt h(|x|) \lt 1\). In terms of balls these are exactly the three asserted regions, completing the proof.

The proof proceeds in five movements: refine the cover, build a bump on each piece, sum them, normalize, and reindex. We write it for a manifold without boundary. At boundary points the regular coordinate balls of Steps 1 and 2 are replaced by regular coordinate half-balls, supplied by the countable basis of regular coordinate balls and half-balls. The open sets \(X_\alpha\) that Step 1 passes to are correspondingly open submanifolds with boundary rather than open submanifolds, and the basis each of them inherits consists of regular coordinate balls and half-balls. The locally finite refinement is then produced by the same argument that gives paracompactness with boundary. Nothing else in the construction changes, so we do not write the boundary case out separately.

Step 1: A refinement by regular coordinate balls.
Each \(X_\alpha\), being open in \(M\), is itself a smooth manifold as an open submanifold, 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 \(M\). Given an open subset \(W \subseteq M\) and a point \(q \in W\), the point \(q\) lies in some \(X_\alpha\), because the \(X_\alpha\) cover \(M\). The intersection \(W \cap X_\alpha\) is open in \(X_\alpha\) and contains \(q\), so \(\mathcal{B}_\alpha\), being a basis for \(X_\alpha\), supplies a regular coordinate ball \(B\) with \(q \in B \subseteq W \cap X_\alpha \subseteq W\). 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 \(B_i\) belongs to \(\mathcal{B}_\alpha\) for at least one \(\alpha\). 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 choose an index \(\alpha\) with \(B_i \in \mathcal{B}_\alpha\). Since \(B_i\) is then 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, \[ \varphi_i(\overline{B_i}) = \overline{B_{r_i}(0)}, \quad \varphi_i(B_i') = B_{r_i'}(0), \quad r_i \lt r_i'. \] Here \(\overline{B_i}\) may be read as the closure in \(X_\alpha\) or as the closure in \(M\) without ambiguity. The map \(\varphi_i\) carries \(\overline{B_i}\) homeomorphically onto the compact set \(\overline{B_{r_i}(0)}\), so \(\overline{B_i}\) is compact. A compact subset of a Hausdorff space is closed, so \(\overline{B_i}\) is closed in \(M\). A closed subset of \(M\) containing \(B_i\) contains the closure of \(B_i\) in \(M\), and the closure in a subspace is always contained in the closure in the ambient space, so the two agree. 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 \(B_{r_i}(0)\). Because \(H_i\) vanishes there, \(H_i \circ \varphi_i = 0\) 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 \(f_i \gt 0\). Hence \(f(p) \gt 0\) 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 \(A\).
Each \(B_i'\) was chosen inside some \(X_\alpha\) in Step 2, so \(\{B_i'\}\) refines \(\mathcal{X}\). For each \(i\) we may therefore 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 \(g_i\) are nonnegative, so \(\psi_\alpha\) is nonzero exactly on \(\bigcup_{i \,:\, a(i) = \alpha} B_i\), and the closure lemma permits the closure of a locally finite union to be computed termwise: \[ \begin{align*} \operatorname{supp} \psi_\alpha &= \overline{\bigcup_{i \,:\, a(i) = \alpha} B_i} \\\\ &= \bigcup_{i \,:\, a(i) = \alpha} \overline{B_i} \subseteq X_\alpha, \end{align*} \] 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 \(\operatorname{supp} \psi_0 \subseteq M \setminus A\). 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, furnished by the basis of precompact coordinate balls when the boundary is empty and by the corresponding basis of precompact coordinate balls and half-balls otherwise. Either 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.

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 \(j \gt N\), 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 \(\sum_{j \gt N} \psi_j(p) = 1\), and therefore \[ f(p) = \sum_j j\, \psi_j(p) = \sum_{j \gt N} j\, \psi_j(p) \ge (N + 1) \sum_{j \gt N} \psi_j(p) = N + 1 \gt N \ge c, \] 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. Hence \(f^{-1}\bigl((-\infty, c]\bigr)\) is compact. Since \(c\) was arbitrary, \(f\) is a smooth exhaustion function.