HW-5: Smooth Bump Functions

We use the standard fact that the function \[ \phi(t)= \begin{cases} e^{-1/t^2}, & t\neq 0,\\ 0, & t=0, \end{cases} \] belongs to \(C^\infty(\mathbb{R})\) and satisfies \(\phi^{(k)}(0)=0\) for every \(k\geq 0\). In other words, \(\phi\) and all its derivatives vanish at \(t=0\).

For \(x\in(-1,1)\), \(f(x) = e^{-1/(x-1)^2}e^{-1/(x+1)^2}\). Both factors are smooth on \((-1,1)\), so their product is also \(C^\infty\) on \((-1,1)\).

For \(x\notin[-1,1]\), the function \(f\) is identically zero and hence is clearly \(C^\infty\) there. It remains only to check smoothness at \(x=1\) and \(x=-1\).

Near \(x=1\), we can write \(f(x) = e^{-1/(x-1)^2}\,e^{-1/(x+1)^2}\). The second factor is a smooth function in a neighborhood of \(x=1\), while the first factor is of the form \(\phi(x-1)\). Since \(\phi^{(k)}(0)=0\) for every \(k\geq0\), all derivatives of \(f\) from the left tend to \(0\) as \(x\to1\). On the other hand, \(f\) is identically zero for \(x\geq1\), so all right-hand derivatives at \(x=1\) are also zero. Hence \(f\) is \(C^\infty\) at \(x=1\).

Similarly, near \(x=-1\), \(f(x) = e^{-1/(x-1)^2}\,e^{-1/(x+1)^2}\), where the first factor is smooth near \(x=-1\) and the second factor is of the form \(\phi(x+1)\). Thus all derivatives of \(f\) tend to zero as \(x\to-1\), and they agree with the derivatives of the zero function outside \((-1,1)\). Therefore \(f\) is \(C^\infty\) at \(x=-1\) as well.

Consequently, \(f\in C^\infty(\mathbb{R}).\)

Finally, for \(x\in(-1,1)\) both exponential factors are strictly positive. Hence \(f(x)>0\) for \(x\in(-1,1)\). By definition, \(f(x)=0\) for \(x\notin(-1,1)\).

Therefore, \(f\) is \(C^\infty\), \(f>0\) on \((-1,1)\), and \(f=0\) outside \((-1,1)\). ∎

From part (a), after a translation and scaling, we can construct a \(C^\infty\) function \(\widetilde f:\mathbb{R}\to[0,\infty)\) which is positive on \((0,\varepsilon)\) and zero outside \([0,\varepsilon]\).

For example, define \(\widetilde f(x) = f\left(\frac{2x-\varepsilon}{\varepsilon}\right)\). Indeed, \(-1 < \frac{2x-\varepsilon}{\varepsilon} < 1\) is equivalent to \(00\) on \((0,\varepsilon)\) and \(\widetilde f=0\) elsewhere.

Since \(\widetilde f\) is positive on \((0,\varepsilon)\), \(\int_0^\varepsilon \widetilde f(t)\,dt>0\).

Define: \[ g(x) = \frac{\displaystyle\int_0^x\widetilde f(t)\,dt}{\displaystyle\int_0^\varepsilon\widetilde f(t)\,dt} \]

Since \(\widetilde f\in C^\infty(\mathbb{R})\), the Fundamental Theorem of Calculus gives \(\frac{d}{dx}\left(\int_0^x\widetilde f(t)\,dt\right) = \widetilde f(x)\). Therefore \(\int_0^x\widetilde f(t)\,dt\) is \(C^\infty\), and hence \(g\in C^\infty(\mathbb{R})\).

If \(x\leq0\), then \(\widetilde f(t)=0\) on the interval between \(x\) and \(0\). Hence \(\int_0^x\widetilde f(t)\,dt=0\), and therefore \(g(x)=0\).

If \(0

Finally, if \(x\geq\varepsilon\), then \(\widetilde f(t)=0\) for \(t\geq\varepsilon\). Therefore, \(\int_0^x\widetilde f(t)\,dt = \int_0^\varepsilon\widetilde f(t)\,dt\), and hence \(g(x)=1\).

Thus, \(g:\mathbb{R}\to[0,1]\) is \(C^\infty\) and satisfies the conditions. ∎

Let \(a=(a^1,\ldots,a^n)\in\mathbb{R}^n\) and define \(g(x) = \prod_{i=1}^n f\left(\frac{x^i-a^i}{\varepsilon}\right)\).

For each \(i\), the function \(x\longmapsto \frac{x^i-a^i}{\varepsilon}\) is a smooth function on \(\mathbb{R}^n\). Since \(f\in C^\infty(\mathbb{R})\), the composition \(x\longmapsto f\left(\frac{x^i-a^i}{\varepsilon}\right)\) is \(C^\infty\) on \(\mathbb{R}^n\).

Since \(g\) is a finite product of \(C^\infty\) functions, it follows that \(g\in C^\infty(\mathbb{R}^n)\).

By part (a), \(f(y)>0 \iff -10\) if and only if \(-1 < \frac{x^i-a^i}{\varepsilon} < 1\). Since \(\varepsilon>0\), this is equivalent to \(a^i-\varepsilon

Thus, all factors in the product are positive precisely when \(a^i-\varepsilon0\) on \((a^1-\varepsilon,a^1+\varepsilon) \times\cdots\times (a^n-\varepsilon,a^n+\varepsilon)\).

If \(x\) lies outside this rectangle, then for at least one index \(i\), \(|x^i-a^i|\geq\varepsilon\). Therefore, \(\left| \frac{x^i-a^i}{\varepsilon} \right|\geq1\), and consequently \(f\left(\frac{x^i-a^i}{\varepsilon}\right)=0\). Thus the whole product is zero: \(g(x)=0\).

Therefore, \(g>0\) on the open rectangle, and \(g=0\) elsewhere. ∎

Let \(A\subset\mathbb{R}^n\) be open and let \(C\subset A\) be compact. If \(C=\varnothing\), the result is immediate by taking \(f\equiv0\). Assume therefore that \(C\neq\varnothing\).

For every point \(a=(a^1,\ldots,a^n)\in C\), since \(A\) is open and \(a\in A\), there exists \(\varepsilon_a>0\) such that the closed rectangle \(Q_a = \prod_{i=1}^n [a^i-\varepsilon_a,a^i+\varepsilon_a]\) is contained in \(A\).

By part (c), there exists a non-negative \(C^\infty\) function \(g_a:\mathbb{R}^n\to\mathbb{R}\) such that \(g_a(x)>0\) on the interior \(Q_a^\circ = \prod_{i=1}^n (a^i-\varepsilon_a,a^i+\varepsilon_a)\) and \(g_a(x)=0\) outside \(Q_a\).

The collection \(\{Q_a^\circ:a\in C\}\) is an open cover of \(C\). Since \(C\) is compact, there exist finitely many points \(a_1,\ldots,a_m\in C\) such that \(C\subset Q_{a_1}^\circ\cup\cdots\cup Q_{a_m}^\circ\).

Let \(g_1,\ldots,g_m\) be the corresponding functions from part (c) and define: \[ f(x)=\sum_{j=1}^m g_j(x), \qquad x\in A. \]

Each \(g_j\) is non-negative and \(C^\infty\). Hence \(f\) is also non-negative and \(C^\infty\).

Now let \(x\in C\). Since the interiors of the rectangles form a finite cover of \(C\), there is some \(j\) such that \(x\in Q_{a_j}^\circ\). Therefore, \(g_j(x)>0\). Since all the other terms are non-negative, \(f(x) = \sum_{k=1}^m g_k(x) \geq g_j(x)>0\). Thus, \(f(x)>0\) for every \(x\in C.\)

Finally, let \(K=Q_{a_1}\cup\cdots\cup Q_{a_m}\). Each \(Q_{a_j}\) is closed, and there are only finitely many of them. Therefore \(K\) is closed. Moreover, \(K\subset A\). Since each \(g_j\) is zero outside \(Q_{a_j}\), we have \(f(x)=0\) for \(x\notin K\).

Hence there exists a non-negative \(C^\infty\) function \(f:A\to\mathbb{R}\) such that \(f(x)>0\) on \(C\), and \(f(x)=0\) outside a closed set \(K\subset A\). ∎

Let \(f\) be the function obtained in part (d). Since \(f(x)>0\) for all \(x\in C\), and \(f\) is continuous, the compactness of \(C\) implies that \(f\) attains its minimum on \(C\). Hence there exists \(\varepsilon>0\) such that \(f(x)\geq\varepsilon\) for every \(x\in C\).

Now let \(g:\mathbb{R}\to[0,1]\) be the \(C^\infty\) function constructed in part (b), corresponding to this value of \(\varepsilon\). Thus, \(g(t)=0\) for \(t\leq0\), and \(g(t)=1\) for \(t\geq\varepsilon\).

Define: \[ F(x)=g(f(x)), \qquad x\in A. \]

Since \(f\in C^\infty(A)\) and \(g\in C^\infty(\mathbb{R})\), their composition is \(C^\infty\). Therefore, \(F\in C^\infty(A)\).

For every \(x\in A\), \(0\leq g(f(x))\leq1\). Hence \(F:A\to[0,1]\).

If \(x\in C\), then \(f(x)\geq\varepsilon\). Therefore, by the defining property of \(g\), \(F(x)=g(f(x))=1\). Thus, \(F(x)=1\) for every \(x\in C\).

Finally, let \(K\subset A\) be the closed set from part (d), outside of which \(f=0\). If \(x\notin K\), then \(f(x)=0\). Since \(g(0)=0\), \(F(x)=g(f(x))=g(0)=0\). Thus \(F\) also vanishes outside the same closed set \(K\subset A\).

Therefore we have constructed a function \(F\in C^\infty(A)\), \(0\leq F\leq1\), \(F=1\) on \(C\), which is zero outside a closed set contained in \(A\). ∎