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Consider a Hilbert-Schmidt operator $A$ on $L^2(\mathbb R^d)$ with integral kernel $A(x,y)$. Let $\Omega\subset \mathbb R^d$ and $1_{\Omega}(x)$ denote its characteristic function as well as the corresponding multiplication operator. Note that by the triangle inequality $$\lVert A\rVert_2 \leq \lVert 1_{\Omega} A \rVert_2 + \lVert 1_{\Omega^c} A \rVert_2,$$ where $\lVert \cdot \rVert_2$ is the Hilbert-Schmidt norm. By the nice property that we can express $\lVert A \rVert_2$ as the $L^2 \times L^2$ norm of its kernel we also have the (in-)equality $$\begin{align} \lVert A \rVert_2^2 & = \int_{\mathbb R^d} \int_{\mathbb R^d} \lvert A(x,y)\rvert^2 \, dx \, dy \\ & = \int_{\Omega} \int_{\mathbb R^d} \lvert A(x,y)\rvert^2 \, dx \, dy + \int_{\Omega^c} \int_{\mathbb R^d} \lvert A(x,y)\rvert^2 \, dx \, dy \\ & = \lVert 1_\Omega A \rVert_2^2 + \lVert 1_{\Omega^c} A \rVert_2^2 \end{align}$$ with squares on both sides. My question: Does this generalize to higher Schatten $p$-norms, $p>2$? That is, does $$\lVert A \rVert_p^2 \leq \lVert 1_\Omega A\rVert_p^2 + \lVert 1_{\Omega^c} A\rVert_p^2$$ hold? If not, does anyone have a good counterexample? Thanks in advance!

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    $\begingroup$ Yes, it is the triangle inequality for the Schatten $p/2$-norm. $\endgroup$ Commented Jun 11, 2021 at 13:38
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    $\begingroup$ Thanks, I think I see it now. More generally, for $A, B$ such that $A^*B = B^*A = 0$ we have $\lVert A + B \rVert_p^2 \leq \lVert A \rVert_p^2 + \lVert B \rVert_p^2$, right? $\endgroup$
    – user271621
    Commented Jun 11, 2021 at 13:42
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    $\begingroup$ That is right, yes. $\endgroup$ Commented Jun 11, 2021 at 14:16

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