A function $f$ on $\mathbb{R}$ is called locally integrable if $f$ is integrable on every bounded interval $[a, b]$ for $a < b$ in $\mathbb{R}$. Show that if $g \in C_c^{\infty}(\mathbb{R})$ and $f$ is locally integrable, then $$f∗g(y)=\int_{-\infty}^{\infty}f(t)g(y-t)dt$$ exists and is infinitely differentiable on $\mathbb{R}$.

I feel pretty much stuck from the beginning. After taking modulus using triangle inequality I was trying to bound things individually but it didn't work out.

Please help.

  • 3
    $\begingroup$ g has compact support (and so it is bounded). this allows you to easily see that the convolution is well defined since for is locally integrable. to show that this is infinitely differentiable look at the difference quotient and see if you can pass the limit through the integral (pass the derivatives onto get instead of f!) $\endgroup$ – yousuf soliman Jul 22 '18 at 7:12
  • $\begingroup$ I missed the subscript $c$ in $C_c^\infty(\Bbb R)$. $\endgroup$ – Kenny Lau Jul 22 '18 at 7:13
  • $\begingroup$ @yousufsoliman The subscript means it's having a compact support? $\endgroup$ – rationalpi Jul 22 '18 at 7:24
  • $\begingroup$ @mathjack51 that is correct $\endgroup$ – yousuf soliman Jul 22 '18 at 7:25

The comment of yousuf soliman is exactly right. Since I've been wanting to write the details of this computation for myself, I'll give them here. Just for some context, some typical examples of unbounded, locally (but not globally) integrable $f$'s are $f(y) = |y|^{-p}$ for $0< p< 1$ and $f(y) = |\log |y||$; the singularities of these functions at the origin are mild enough that they can be integrated over.

Here are the details as outlined by yousef. Fix $g\in C_c^\infty(\mathbb R)$ and choose $R>0$ such that the support of $g$ is contained in $[-R, R]$. For each $y\in \mathbb R$ the support of the function $t\mapsto g(y - t)$ is contained in $[y - R, y+ R]$. First, I'll verify that the convolution is pointwise well-defined on $\mathbb R$. For $y\in \mathbb R$ fixed, we have \begin{eqnarray*} \left|f*g(y)\right| & = & \left|\int_{\mathbb R} f(t) g(y - t)\; {\rm d} t\right|\\ & = & \left|\int_{y - R}^{y + R} f(t) g(y - t)\; {\rm d} t\right|\\ & \leq & \int_{y - R}^{y + R} |f(t)|\; |g(y - t)|\; {\rm d} t\\ & \leq & \max_{x\in \mathbb R}|g(x)|\int_{y - R}^{y + R} |f(t)|\; {\rm d}t\\ & < & \infty. \end{eqnarray*}

Next, I'll use the Dominated Convergence Theorem to show that $f*g$ is differentiable and that the derivative of $f*g$ is equal to the convolution of $f$ with $g'$. With $R$ as above, let $0< |h|< \frac{R}{100}$ so that for each $y\in \mathbb R$, the supports of both of the functions $t\mapsto g(y - t)$ and $t\mapsto g(y + h- t)$ are contained in the interval $[y - 2R, y + 2R]$. Then \begin{eqnarray*} \frac{f*g(y + h) - f*g(y)}{h} & = & \int_{\mathbb R}f(t)\frac{g(y + h- t) - g(y- t)}{h}\; {\rm d}t \\ & = & \int_{y - 2R}^{y + 2R}f(t)\frac{g(y + h- t) - g(y- t)}{h}\; {\rm d}t. \end{eqnarray*} Since $f$ is integrable on the interval $[y - 2R, y+ 2R]$, we have $|f(t)|< \infty$ for a.e. $t\in [y - 2R, y + 2R]$. Therefore, for a.e. $t\in [y - 2R, y + 2R]$ we have \begin{equation*} \lim_{h\to 0}f(t)\frac{g(y + h- t) - g(y- t)}{h} = f(t) g'(y - t). \end{equation*} It remains to find a function $\phi(t)$, integrable over $[y- 2R, y + 2R]$ for which \begin{equation*} \left|f(t)\frac{g(y + h - t) - g(y- t)}{h}\right| \leq \phi(t) \end{equation*} for a.e. $t\in [y - 2R, y+ 2R]$. By the Mean Value Theorem we have \begin{eqnarray*} \left|g(y + h - t) - g(y - t)\right| & = & \left| \int_0^1\frac{\rm d}{{\rm d}s} g(x - t + sh)\; {\rm d} s \right| \\ &\leq & |h|\max_{x\in \mathbb R} |g'(x)|. \end{eqnarray*} As a consequence, for all $0< |h| < \frac{R}{100}$, \begin{equation*} \left|f(t) \frac{g(y + h -t) - g(y - t)}{h}\right| \leq (\max_{x\in \mathbb R} |g'(x)|)|f(t)| \end{equation*} with the function on the right-hand side integrable over $[y - 2R, y+ 2R]$. Finally, the Dominated Convergence Theorem gives \begin{eqnarray*} \lim_{h\to 0}\frac{f*g(y + h) - f*g(y)}{h} & = & \int_{y - 2R}^{y + 2R} \lim_{h\to 0} f(t) \frac{g(y + h - t) - g(y - t)}{h} \; {\rm d}t \\ & = & \int_{y - 2R}^{y + 2R} f(t)g'(y - t)\; {\rm d}t \\ & = & \int_{\mathbb R} f(t) g'(y - t)\; {\rm d}t \\ & = & f*g'(y). \end{eqnarray*}

Finally, to see that $f*g$ is infinitely differentiable, one may proceed by induction on the order of differentiability of $f*g$. If $f*g$ is $k$-times differentiable with $(f*g)^{(k)} = f*(g^{(k)})$ then apply the above argument with $g$ replaced by $g^{(k)}$ to find that $f*g$ is $(k + 1)$-times differentiable with $(f*g)^{(k+1)} = f*(g^{(k+1)})$.


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