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

Question

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 $W^{1,\infty}$?

Answer 1

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*}

Answer 2

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$$(K*f)(x)=\int K(y)f(x-y)dy=\int\int\mathbf{1}(z<y) f(x-y)dm(z) dy\\= \int dm(z) \int_z^\infty f(x-y)dy=\int dm(z)\int_{-\infty}^{x}f(t-z)dt=\int_{-\infty}^xdt\int f(t-z)dm(z)$$ is an "antiderivative" of a bounded $L^1$ function $x\to \int f(x-z)dm(z)$.