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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.

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  • $\begingroup$ Such users are both discourteous and annoying. $\endgroup$ – Bill Johnson Mar 17 '18 at 23:13
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The answer is no. Indeed, let \begin{equation} k(x,y):=f(x)g(y), \end{equation} 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 \begin{equation} \|K\|_{\infty,\infty}=\sup_x\int|k(x,y)|dy=\|f\|_\infty \|g\|_1=\infty. \end{equation} So, the inequality $\|K\|_{\infty,\infty}\le c\|K\|_{tr}$ is false for any real $c$.

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