The convolution off and $g$ on $\mathbb{R}^d$ equipped with Lebesgue measure is defined by $$f*g(x) = \int_{\mathbb{R}^d} f(x-y)g(y)dy.$$ Define $||f||_{L^\infty} =$ inf$\{ M : |f(x)|< M$ for $\mu$ a.e. $x \in X\}$.
Assume $\frac{1}{p} + \frac{1}{q} = 1$ and $f \in L^p, g \in L^q$. Prove $f*g \in L^\infty$ and $||f*g||_{L^\infty} \leq ||f||_{L^p}||g||_{L^q}$.
My fist attempt:
Let $|g(y)|dy = d\mu_2$. Then $||f*g||_\infty \leq || \,|f|*|g| \.||_{L^\infty} = || \int_{\mathbb{R}^d } |f(x-y)| \, |g(y)| \, dy ||_{L^\infty}$
$= || \int_{\mathbb{R}^d} |f(x-y)| d \mu_2 ||_{L^\infty} \leq \int_{\mathbb{R}^d} || \,|f(x-y)| \, ||_{L^\infty} d \mu_2 = \int_{\mathbb{R}^d} || \,|f(x)| \, ||_{L^\infty} d \mu_2$
$= \| |f(x)| \|_{L^\infty} \int_{\mathbb{R}^d} = \| |f(x)| \|_{L^\infty} \int_{\mathbb{R}^d} |g(y)| dy = \| |f(x)| \|_{L^\infty} \|g\|_1$
But I don't think $\| |f(x)| \|_{L^\infty} \|g\|_1 \leq ||f||_{L^p}||g||_{L^q}$.
My second attempt:
So for any $\epsilon > 0$, since $||f||_{L^\infty} - \epsilon < ||f||_{L^\infty} =$ inf$\{ M : |f(x)|< M$ for $\mu$ a.e. $x \in X\}$, it must be the case that $\{x : |f(x)| > ||f||_\infty - \epsilon\}$ has measure greater then $0$. In other words for every $\epsilon > 0$ there exists a $\delta > 0$ such that $\mu(\{x : |f(x)| > ||f||_\infty - \epsilon\}) > \delta$.
Let $|g(y)|dy = d\mu_2~$, we get $$||f*g||_{L^\infty} = ||\int_{\mathbb{R}^d} f(x - y)g(y)dy||_{L^\infty} = ||\int_{\mathbb{R}^d} f(x - y)d\mu_2|| _{L^\infty} \leq \int_{\mathbb{R}^d} ||f(x - y)||_{L^\infty} d\mu_2 = \int_{\{x : |f(x)| > ||f||_{L^\infty} + \epsilon\}} ||f||_{L^\infty} d\mu_2 + \int_{\{x : |f(x)| \leq |f||_{L^\infty} + \epsilon\}} ||f||_{L^\infty} d\mu_2~.$$
But I feel like this is not the right approach as I now have the sum of two integrals instead of the product of integrals.
My third attempt:
By Holdors inequality $\|f\|_{L^p}\|g\|_{L^q} \geq \|fg\|_{L_1}$
But I don't think $\|fg\|_{L_1} \geq ||f * g||_{L\infty}$.
I think this might be an answer
$|f*g| = |\int_{\mathbb{R}^d} f(x - y)g(y)dy| \leq \int_{\mathbb{R}^d} |f(x- y)g(y)|dy = \|f(x - y)g(y)\|_{L^1} \leq \|f(x - y)\|_{L^p}\|g\|_{L^q} = \|f\|_{L^p}\|g\|_{L^q}$ where the fourth inequality follows from Holders inequality and the fifth inequality holds by invariants of the Lebesgue measure. Therefore $\|f\|_{L^p}\|g\|_{L^q}$ is an upper bound for $|f*g|$, for almost every $x$. So since $\|f*g\|_{L^\infty}$ is the inf $|f*g|$ for almost every $x$ we have $\|f*g\|_{L^\infty} \leq \|f\|_{L^p}\|g\|_{L^q}$