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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 $W^{1,\infty}$?

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2 Answers 2

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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*}

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

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  • $\begingroup$ 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 $x$. $\endgroup$ Commented Jan 9, 2021 at 12:34
  • $\begingroup$ Thank you. I put the quesiton on the second derivative in a separate post: mathoverflow.net/questions/380772/… $\endgroup$
    – Hiro
    Commented Jan 9, 2021 at 13:24

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