Let $p, q \in [1, \infty]$ such that $\frac{1}{p}+\frac{1}{q} = 1$. We define the convolution operator $$ * : L^{p} (\mathbb R^d) \times L^{q} (\mathbb R^d) \to L^\infty (\mathbb R^d) $$ by $$ (f*g) (x) := \int f(x-y) g(y)dy \quad \forall x \in \mathbb R^d. $$
By Hölder's inequality, $$ \begin{align} |(f*g) (x)| &\le \bigg [ \int |f(x-y)|^p dy \bigg ]^{1/p} \bigg [ \int |g(y)|^q dy \bigg ]^{1/q} \\ &= \|f\|_p \|g\|_q < \infty \quad \forall x \in \mathbb R^d. \end{align} $$
Hence $$ \|f*g\|_\infty \le \|f\|_p \|g\|_q. $$
So the symmetric bi-linear map $*$ is continuous. I'm trying to prove a result mentioned in this comment, i.e.,
Theorem $f*g$ is uniformly continuous.
Could you have a check on my below attempt?
Proof Let $x_n, y_n \in \mathbb R^d$ such that $h_n := x_n-y_n \to 0$. Clearly, at least one element of $\{p, q\}$ is finite. WLOG, we assume $p < \infty$. Then $$ \begin{align*} & |(f*g)(x_n) - (f*g)(y_n) | \\ \le{} & \int |f(x_n-y) - f(y_n-y)| \cdot |g(y)|dy \\ \le{} &\|g\|_q \bigg [ \int |f(x_n-y) - f(y_n-y)|^p dy \bigg ]^{1/p} \quad \text{by Hölder's inequality}\\ \le{} &\|g\|_q \bigg [ \int |f(y+h_n) - f(y)|^p dy \bigg ]^{1/p} \quad \text{by translation-invariance of } dy. \end{align*} $$
The claim then follows from below lemma, i.e.,
Lemma Let $f \in L^p(\mathbb{R^d})$ with $p \in [1, \infty)$. We define the shift operator $\tau_h$ by $(\tau_h f)(x) := f(x−h)$ for all $h, x \in \mathbb R^d$. Then $\lim_{h \to 0} \|\tau_h f−f\|_{p} =0$, i.e., $$ \lim_{h\to 0} \left [ \int_{\mathbb R^d} |f(x-h) - f(x)|^pd x\right]^{1/p} =0. $$
Sure, or the more succinct way of phrasing the argument is that \begin{align} \sup_{\Bbb{R}^n}|\tau_h(f*g)-f*g|&=\sup_{\Bbb{R}^n}|(\tau_hf-f)*g|\leq\|\tau_hf-f\|_{L^p}\|g\|_{L^{q}}, \end{align} and indeed without loss of generality, you can assume $p$ is finite, so by continuity of translations (more precisely, in the strong-operator topology), the RHS vanishes as $h\to 0$. This is equivalent to uniform continuity of $f*g$.