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user271621
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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 $$\lVert A \rVert_p^2 \leq \lVert 1_\Omega A\rVert_p^2 + \lVert 1_{\Omega} A\rVert_p^2$$$$\lVert A \rVert_p^2 \leq \lVert 1_\Omega A\rVert_p^2 + \lVert 1_{\Omega^c} A\rVert_p^2$$ hold? If not, does anyone have a good counterexample? Thanks in advance!

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 $$\lVert A \rVert_p^2 \leq \lVert 1_\Omega A\rVert_p^2 + \lVert 1_{\Omega} A\rVert_p^2$$ hold? If not, does anyone have a good counterexample? Thanks in advance!

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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user271621
user271621
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Additivity of squared Schatten $p$-norm with respect to spatial partition

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 $$\lVert A \rVert_p^2 \leq \lVert 1_\Omega A\rVert_p^2 + \lVert 1_{\Omega} A\rVert_p^2$$ hold? If not, does anyone have a good counterexample? Thanks in advance!