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$\phi \in L^1(\mathbf R)$ where $L^1(\mathbf R) = \{ \phi: \mathbf R \to \mathbf R$ such that $\int_\mathbf R |\phi(\mathbf x)|d\mathbf x < \infty\}$

and $f \in C_c(\mathbf R)$ where $C_c(\mathbf R) = \{f \in C(\mathbf R): supp(f)\}$

where $supp(f) := \{x:f(\mathbf x) \neq 0\}$

and I am asked to show that if this is true then the convolution $\phi * f \in L^\infty(\mathbf R)$

$(\phi * f)(\mathbf x) = \int_\mathbf R \phi(\mathbf x - \mathbf y)f(\mathbf y)d\mathbf y$

My attempt: So as $\phi \in L^1(\mathbf R)$ then I know it is bounded and closed. So I can say $\exists M >0$ such that $\phi(\mathbf x - \mathbf y)<M$

So I can write the integral $\int_\mathbf R \phi(\mathbf x - \mathbf y)f(\mathbf y)d\mathbf y < M \int_\mathbf R f(\mathbf y) d\mathbf y$, but I don't know how to go on from here to show that this is $\in L^\infty$

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This is a special case of Young's convolution inequality and doesn't require that $\phi$ be continuous with compact support, but merely essentially bounded.

Given $x$ we have

$|f\ast \phi (x)| = | \int_{\Bbb R}f(x -y)\phi(y) dy| \leq \int_{\Bbb R}|f(x - y)|||\phi ||_{\infty} dy = ||f||_1 ||\phi||_{\infty} $

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You have $$|\varphi*f(x)| \le \int |\varphi(\xi)f(x - \xi)|\, d\xi \le \int |\varphi(\xi)|\|f\|_\infty\, d\xi = \|\varphi_1\|\|f\|_\infty$$

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