Let $k\in C(\mathbb{R}^2; \mathbb{R})$ be a continuous function. Suppose that the operator $K\colon L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ defined by the formula $$(Kf)(x)=\int_{\mathbb{R}} k(x,y)f(y) dy$$ has a finite trace norm $\|K\|_{tr}$. Can we bound the operator norm $\|K\|_{\infty,\infty}$ of $K$ considered as an operator from $L^{\infty}(\mathbb{R})$ to $L^{\infty}(\mathbb{R})$ by $c\|K\|_{tr}$, where $c$ is a finite universal constant?

This question was asked by a user and then deleted by the OP while I was preparing the answer. So, I am going to give the answer below.