Can the $L^{\infty}\to L^{\infty}$ norm be bounded by the trace norm?

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.

• Such users are both discourteous and annoying. – Bill Johnson Mar 17 '18 at 23:13
• I agree, there are some ups and downs here. Overall, my MathOverflow experience has been useful and exciting to me. I have learned quite a bit from a number of users, including yourself. – Iosif Pinelis Mar 18 '18 at 12:50

The answer is no. Indeed, let $$k(x,y):=f(x)g(y),$$ where $f$ and $g$ are any continuous functions in $L^2(\mathbb{R})$ such that $f\ne0$ and $\|g\|_1=\infty$; e.g., one may take $f(x)=g(x)=\frac1{\sqrt{1+x^2}}$. Then $K^*Kh=\|f\|_2^2\,(\int gh) g$ for $h\in L^2(\mathbb{R})$ and hence $\|K\|_{tr}=\|f\|_2\|g\|_2<\infty$, whereas $$\|K\|_{\infty,\infty}=\sup_x\int|k(x,y)|dy=\|f\|_\infty \|g\|_1=\infty.$$ So, the inequality $\|K\|_{\infty,\infty}\le c\|K\|_{tr}$ is false for any real $c$.