Estimate for convolution

Question

Let $f \in L^p(\mathbb R)$ for $1 \leq p < \infty$, $\varphi \in L^1(\mathbb R)$ and $A \subseteq \mathbb R$ Borel-measurable with $\lambda(A) < \infty$. I want to show that $$\int_{\mathbb R} \int_{\mathbb R} \vert \varphi(y) f(x-y) \chi_A(x)\vert\,dy\,dx < \infty \qquad (1)$$ to deduce that $g_x \in L^1(\mathbb R)$, $g_x(y) = \varphi(y) f(x-y)$ for $x \in \mathbb R$ almost everywhere.

But I don't know how to do that. I tried to use Fubini to get

\begin{align*} \int_{\mathbb R} \int_{\mathbb R} \vert \varphi(y) f(x-y) \chi_A(x)\vert\,dy\,dx &= \int_{\mathbb R} \vert \varphi(y) \vert \int_{\mathbb R} \vert f(x-y) \chi_A(x)\vert\,dx\,dy \\ &= \int_{\mathbb R} \vert \varphi(y) \vert \int_{A} \vert f(x-y) \vert\,dx\,dy \\ &= \int_{\mathbb R} \vert \varphi(y) \vert \int_{A - y} \vert f(x) \vert\,dx\,dy \end{align*} Now I would like to bound $\int_{A - y} \vert f(x) \vert\,dx$ by a constant but I don't see how this works. Another idea of me was to approximate $f$ by elements from $C^\infty_c(\mathbb R)$, because I can show that $(1)$ holds for $f \in C^\infty_c(\mathbb R)$ but I can't deduce $(1)$ for $f \in L^p(\mathbb R)$ from that.

Also I have no idea how to deduce $g_x \in L^1(\mathbb R)$ from $(1)$. I would appreciate some hints :)

Answer 1

Simply use Hölder's inequality to derive \begin{align*} \int_{A-y} |f(x)| dx &= \int |f(x)| \cdot 1_{A-y} (x) dx \\ &\leq \| f \|_{L^p} \cdot \| 1_{A-y} \|_{L^{p'}} \\ &= \| f \|_{L^p} \cdot \| 1_{A} \|_{L^{p'}}, \end{align*} where the right-hand side is independent of $y$.

To deduce that $g_x \in L^1$ for almost all $x $, note that the expression in (1) can be written as $$ \int \| g_x \|_{L^1} \cdot 1_A (x) dx. $$ Now, shot that if $\int h (x) 1_A (x) dx < \infty $ for some $h \geq 0$ and ever set $A $ of finite measure, then $h (x) < \infty $ almost everywhere.

Hint: Consider $B = \{x \, : \, h (x)=\infty\} $. It suffices to show that $B $ is a null set.

$ $

Further hint: It suffices to show that each $A_\ell = B \cap [-\ell, \ell] $ is a null set.

Answer 2

You have to use Minkowski's inequality for integrals (see the bottom of this page Minkowski) and Fubini's theorem : \begin{align*} \left\Vert \int_{\mathbb{R}}\left\vert f\left( \cdot-{y}% \right) \varphi\left( {y}\right) \right\vert \,d{y}% \right\Vert _{L^{p}} & \leq\int_{\mathbb{R}}\left\Vert f\left( \cdot-{y}\right) \varphi\left( {y}\right) \right\Vert _{L^{p}}\,d{y}\\ & =\int_{\mathbb{R}}\left\vert \varphi\left( {y}\right) \right\vert \left\Vert f\left( \cdot-{y}\right) \right\Vert _{L^{p}}\,d{y}=\left\Vert f\right\Vert _{L^{p}}\int_{\mathbb{R}}\left\vert \varphi\left( {y}\right) \right\vert \,d{y}, \end{align*} where the last equality follows from the fact that the Lebesgue measure is translation invariant. Since an $L^p$ function is finite $\mathcal{L}^1$ a.e. everywhere you get $\int_{\mathbb{R}}\left\vert f\left( {x}-{y}% \right) \varphi\left( {y}\right) \right\vert \,d{y}<\infty$ for $\mathcal{L}^{1}$-a.e. ${x}\in\mathbb{R}$.