# Prove that $K \ast f \in W^{1,\infty}(\mathbb R)$ if $K \in BV(\mathbb R)$

Let $$f \in L^1 \cap L^\infty(\mathbb R)$$ and $$K \in BV(\mathbb R)$$. Do these assumptions suffice to prove that for the convolution $$K \ast f$$ we have that $$K \ast f \in W^{1,\infty}(\mathbb R)$$ holds?

From this post, I know that this convolution is BV, but is it true that it is also $$W^{1,\infty}$$?

You can also use that for $$g\in L^p(\mathbf{R})$$, \begin{align*} \|\tau_h g-g\|_p\lesssim |h|, \end{align*} characterizes elements $$g$$ of $$W^{1,p}(\mathbf{R})$$ for $$p>1$$ and elements of $$BV(\mathbf{R})$$ for $$p=1$$. In your case you have for $$h\in\mathbf{R}$$, by Young's inequality (which boils down to Hölder's with the exponents involved here) \begin{align*} \|\tau_h (f\star K)-f\star K\|_\infty = \|f\star(\tau_h K-K)\|_\infty \leq \|f\|_\infty \|\tau_h K-K\|_1 \lesssim\|f\|_\infty |h|. \end{align*}

• Thank you! Could you make the Lipschitz constant in the last inequality explicit?
– Hiro
Jan 9, 2021 at 20:08
• Well it should be $|K'|(\mathbf{R})$. Jan 9, 2021 at 21:12
• What does it denote? The total variation of $K$?
– Hiro
Jan 9, 2021 at 22:04
• Yes. If $K\in W^{1,1}(\mathbf{R})$, it would be $\|K'\|_1$ Jan 9, 2021 at 22:19

Without loss of generality $$K$$ is bounded increasing left-continuous, then $$K(x)=m(-\infty,x)$$ for a bounded non-negative measure $$m$$. We have by Fubini theorem$$(K*f)(x)=\int K(y)f(x-y)dy=\int\int\mathbf{1}(z is an "antiderivative" of a bounded $$L^1$$ function $$x\to \int f(x-z)dm(z)$$.

• The function $h(x):=\int f(x-z)dm(z)$ is essentially bounded (since $f$ is bounded and $m$ is a bounded measure). Thus $\int_{-\infty}^x h(t)dt$ is Lipschitz in $x$. Jan 9, 2021 at 12:34
• Thank you. I put the quesiton on the second derivative in a separate post: mathoverflow.net/questions/380772/…
– Hiro
Jan 9, 2021 at 13:24