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Let $f(x) \in L^1 (\mathbb{R})$ and let $g$ be a bounded, continuous and integrable function on $\mathbb{R}$. I want to prove that $F(x) = \int_{\mathbb{R}} f(y) g(xy) dy$ is continuous.

By definition, I need to check that $|F(x_1) - F(x_2)|$ is small enough if $|x_1-x_2|$ is small enough. I have problems with estimating $|F(x_1)-F(x_2)| \le \int_{\mathbb{R}}|f(y)||(g(x_1 y) - g(x_2 y))| dy$ since it's hard to estimate $g(x_1 y) - g(x_2 y)$ more clever than $2 \sup_{\mathbb{R}} g$ when $y$ is big enough, so I don't understand how to use continuity of $g(x)$. Any ideas would be appreciated!

iou
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  • I would try to use the idea that $L_1$ functions are "nearly" compactly-supported, in that the tails go to 0. https://math.stackexchange.com/questions/745894/what-does-it-mean-to-be-an-l1-function . So you can break up the integral into compact part in the middle where it has normal normal support and the tails where it goes to 0, then in the middle you can't have $y$ spread that far. That might work. – Alan Aug 08 '21 at 21:51

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Fix $x\in\mathbb{R}$. Let $x_n$ be a sequence tending to $x$. For every $y\in\mathbb{R}$, the sequence $f(y)g(x_ny)$ converges to $f(y)g(xy)$, because $g$ is continuous. Moreover, since $g$ is also bounded, we have, for all $n$ and all $y$, $$|f(y)g(x_ny)|\leq M|f(y)|$$ for some $M>0$. Hence, since $f\in L^1(\mathbb{R})$, Lebesgue's dominated convergence theorem can be applied to deduce that $F(x_n)\to F(x)$ as $n\to\infty$.

uniquesolution
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