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Suppose that $\varphi,\gamma\in L^2([0,1],\mathbb C)$ and define the operator $\varphi\otimes\gamma:L^2([0,1],\mathbb C)\to L^2([0,1],\mathbb C)$ by setting $\varphi\otimes\gamma=\langle\cdot,\gamma\rangle\varphi$. This is an integral operator with the kernel given by $\varphi(x)\overline\gamma(y)$ for each $x,y\in [0,1]$. Hence, the Hilbert-Schmidt norm of this operator is given by $$ \|\varphi\otimes\gamma\|_{HS}=\biggl(\int_0^1\int_0^1|\varphi(x)\overline\gamma(y)|^2dxdy\biggr)^{1/2}=\|\varphi\|\|\gamma\|. $$ I want to find the singular values of this operator so that I can obtain the nuclear norm. Since the adjoint of $\varphi\otimes\gamma$ is given by $\gamma\otimes\varphi$, we have that $$ (\varphi\otimes\gamma)^*(\varphi\otimes\gamma)(f)=\langle f,\gamma\rangle\|\varphi\|^2\gamma $$ and if we set $f=\gamma/\|\gamma\|$, we obtain $$ (\varphi\otimes\gamma)^*(\varphi\otimes\gamma)(\gamma\|\gamma\|)=\|\varphi\|^2\|\gamma\|^2\gamma/\|\gamma\|. $$ The composition of operators $(\varphi\otimes\gamma)^*(\varphi\otimes\gamma)$ has one non-zero eigenfunction $\gamma/\|\gamma\|$ with the corresponding eigenvalue $\|\varphi\|^2\|\gamma\|^2$. Hence, the operator $\varphi\otimes\gamma$ has one non-zero singular value given by $\|\varphi\|\|\gamma\|$ and the nuclear norm of $\varphi\otimes\gamma$ is equal to $\|\varphi\|\|\gamma\|$, which is also the Hilbert-Schmidt of this operator.

Are my calculations correct? Does the composition of operators $(\varphi\otimes\gamma)^*(\varphi\otimes\gamma)$ have only one non-zero eigenfunction and the corresponding non-zero eigenvalue? Are the Hilbert-Schmidt and the nuclear norms really equal for this operator?

Any help is much appreciated!

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  • $\begingroup$ Nuclear norm, is that the trace norm in this setting? $\endgroup$ – Jonas Dahlbæk Jun 26 '17 at 15:56
  • $\begingroup$ @JonasDahlbæk It is the Schatten norm with $p=1$ (see here for more details). $\endgroup$ – Cm7F7Bb Jun 27 '17 at 8:29
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Yes. The operators $T=(\varphi\otimes\gamma)^*(\varphi\otimes\gamma) $ you are looking at, are rank one and positive. Concretely, you have that your operator is $$ Tf=\|\varphi\|^2\,\langle f,\gamma\rangle\,\gamma={\|\varphi\|^2}{\|\gamma\|^2}\,\langle f,\gamma'\rangle\,\gamma', $$ where $\gamma'=\gamma/\|\gamma\|$. Written this way, $T$ is a scalar multiple of the projection operator onto the span of $\gamma$. So its spectrum consists of $0$ and $\|\varphi\|^2\|\gamma\|^2$. So, for the HS norm, $$ \|\varphi\otimes\gamma\|_{\rm HS}^2=\text{Tr}\,((\varphi\otimes\gamma)^*(\varphi\otimes\gamma))^{1/2}=(0+\|\varphi\|^2\|\gamma\|^2)^{1/2}=\|\varphi\|\,\|\gamma\|. $$ For the operator norm, $$ \|\varphi\otimes\gamma\|=\|(\varphi\otimes\gamma)^*(\varphi\otimes\gamma)\|^{1/2}=\|\varphi\|\|\gamma\|. $$

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  • $\begingroup$ Thank you very much (+1)! (1) How do we get that $Tf=\|\varphi\|^2\langle f,\gamma\rangle\gamma$? The operator is defined as $\varphi\otimes\gamma=\langle\cdot,\gamma\rangle\varphi$ and I get $Tf=\langle f,\gamma\rangle\varphi$. (2) I think the HS norm does not need to be squared here. (3) For this operator, actually all Schatten norms (for each $1\le p\le \infty$) are equal to its non-zero singular value, right? $\endgroup$ – Cm7F7Bb Jun 27 '17 at 8:48
  • $\begingroup$ What I call $T $ is $(\varphi\otimes\gamma)^*(\varphi\otimes\gamma) $. $\endgroup$ – Martin Argerami Jun 27 '17 at 13:40

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