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The problem:

Suppose $f,g\in L^1(\mathbb{R})$. Let $x\in \mathbb{R}$ and $\phi_x(y) = f(y)g(x-y)$. Show that for almost all $x$, $\phi_x$ is integrable. For such $x$ let $\psi(x) = \int_{-\infty}^\infty \phi_x(y)dy$ and let $\psi(x) = 0$ if $\phi_x$ is not integrable. Show $$ \int_{-\infty}^\infty |\psi(x)|dx \leq \int_{-\infty}^\infty |f(x)|dx\int_{-\infty}^\infty |g(x)|dx.$$

What I've done:

(Thought) To show that the integral of $f(x-y)g(y)$ is finite (i.e., $f(x-y)g(y) \in L^1(\mathbb{R})$) we know H$\ddot{\text{o}}$lder's inequality tells us $$ \| f(y)g(x-y)\|_1 \leq \|f(y)\|_p\|g(x-y)\|_q$$ for all $1\leq p,q\leq \infty$ with $\frac{1}{p} + \frac{1}{q} = 1$. In particular, if we pick $p = 1$ we know $\|f(y)\|_1$ is finite, and as $g\in L^1(\mathbb{R})$ we know it must have finite $L^\infty$ norm, so $\| f(y)g(x-y)\| < \infty$, and hence $f(y)g(x-y)$ is integrable almost everywhere.

Let $E_{y,x} = \{y\in \mathbb{R}: \phi_x \text{ is integrable }\}$. Then by Fubini's theorem we have \begin{align*} \int_{-\infty}^\infty |\psi(x)|dx &= \int_{-\infty}^\infty \left|\int_{E_{y,x}} \phi_x(y) dy\right| dx\newline &= \int_{-\infty}^\infty \left|\int_{E_{y,x}} f(y)g(x-y) dy\right| dx\newline &= \int_{E_{y,x}} \left|\int_{-\infty}^\infty f(y)g(x-y) dx\right| dy\newline &= \int_{E_{y,x}} |f(y)|\left|\int_{-\infty}^\infty g(x-y) dx\right| dy\newline &\leq \int_{E_{y,x}} |f(y)|\int_{-\infty}^\infty |g(x-y)| dx dy\newline &= \int_{E_{y,x}} |f(y)|\int_{-\infty}^\infty |g(x)| dx dy\newline &= \int_{E_{y,x}} |f(y)|\|g\|_1 dy\newline &= \|f\|_1\cdot \|g\|_1, \end{align*} Hence $$ \int_{-\infty}^\infty |\psi(x)|dx \leq \int_{-\infty}^\infty |f(x)|dx\int_{-\infty}^\infty |g(x)|dx.$$

My issue is the area where I claim we can replace $g(x-y)$ with $g(x)$. It's true for any numerical $y$, but inside the integral, with respect to $x$, $y$ is really a variable, so it seems fishy...


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This looks fine for me. The only thing that's missing for me is that you don't explicitly say what you know about $E_{y,x}$ and why (in particular the first equality is somewhat unjustified in your exposition). – t.b. Dec 7 '11 at 5:30
@t.b. Hm.. I guess all I have shown at this point, is that $\psi(x) \in L^1(E_{y,x})$, but $E_{y,x}$ could be really small. I suppose Holder's inequality on $\int f(x-y)g(y)dy$ shows the integral is finite almost everywhere, so that $E_{x,y}$ is almost all of $\mathbb{R}$. – Alex Dec 7 '11 at 5:46
How do you apply Hölder, exactly? Can you include that in your question? – t.b. Dec 7 '11 at 5:49
@t.b. Sure! I suppose it may also be worth actually stating I'm using $p = q = 1$ for H$\ddot{\text{o}}$lder's inequality. – Alex Dec 7 '11 at 6:04
Wait, that won't work as $1/1 + 1/1 \neq 1$... I'll think about this some more... – Alex Dec 7 '11 at 6:08

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