Given a function $f \in L^1$ and a (compactly supported) bounded kernel $k$, this answer suggests to use the Dominated Convergence Theorem to get $$ D(f \star k)(x) = f \star (Dk)(x). $$ My question is: what is the dominating function $g$, which is a necessary condition to be able to apply the DCT?

Additionally, can the conditions be weakened such that $f$ is only locally integrable and $k$ has non-compact support, as long as the convolution integral still exists?

My attempt:

The DCT is used in the following way:

\begin{align} D(f \star k)(x) =& \lim_{n\to\infty}n(f\star k(x+\tfrac1n) - f\star k(x))\\ =&\lim_{n\to\infty} \int n \, f(t) \, (k(x+\tfrac1n-t)-k(x+\tfrac1n)) \, dt \\ {(DCT)? \atop =}& \int \lim_{n\to\infty} n \, f(t) \, (k(x+\tfrac1n-t)-k(x+\tfrac1n)) \, dt \\ =& \int f(t) \, Dk(x-t) \, dt \\ =& f \star (Dk)(x). \end{align}

Naming the terms under the integral \begin{align} h_n &:= n \, f(t) \, (k(x+\tfrac1n-t)-k(x+\tfrac1n)), \\ h &:= f(t) \, Dk(x-t), \end{align} we have pointwise convergence of $h_n \to h$. To apply the DCT, we need to find a dominating function $g$ which satisfies: $$ |h_n(x)| \le g(x). $$

  • $\begingroup$ So $k$ is $C^\infty$? $\endgroup$ – zhw. Nov 9 '15 at 17:44

Because $k$ is smooth with compact support, we have $\sup_{\mathbb {R}}|k'|=M < \infty.$ We are looking at

$$\frac{f\ast k (x+h) - f\ast k (x)}{h} = \int f(t)\frac{k (x+h -t) - k(x-t)}{h}\, dt.$$

By the MVT, the difference quotient of $k$ equals $k'$ evaluated somewhere. Therefore it is bounded uniformly, in absolute value, by $M.$ Hence a dominating function is $|f|\cdot M.$

  • $\begingroup$ I was actually only considering $k \in C^1$. As far as I can see, your answer still applies if I demand k' is bounded, which is implied by compact support. Am I correct? $\endgroup$ – Andre Nov 9 '15 at 21:18
  • 1
    $\begingroup$ Yes, the above works for $C^1.$ (Or even just $k$ is differentiable everywhere with $k'$ bounded.) $\endgroup$ – zhw. Nov 9 '15 at 21:21

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