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I am trying to prove $\|T\| \leq \|T\|_{HS}$ I understand everything up until the following two lines, could somebody please explain why

$\| Tx \| \leq \|T \|_{HS}$ $\|x\|$

implies that

$\| T \| \leq \| T \|_{HS}$

It seems to be so trivial, but I lack the insight.

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  • $\begingroup$ Where is it from? And what means HS? $\endgroup$ – checkmath May 8 '12 at 12:44
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The operator norm is defined as $\sup\left\{\frac{\lVert Tx\rVert}{\lVert x\rVert} ,x\in H, x\neq 0\right\}$. Since by hypothesis each terms of the set $\left\{\frac{\lVert Tx\rVert}{\lVert x\rVert} ,x\in H, x\neq 0\right\}$ is $\leq \lVert T\rVert_{HS}$, we get what we want.

Of course, the most difficult was to prove that $\lVert Tx\rVert\leq \lVert x\rVert\cdot \lVert T\rVert_{HS}$, which need Bessel-Parseval inequality.

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  • $\begingroup$ Seems obvious now, thanks! Yes that was considerably harder, but I can do it, so all is good! $\endgroup$ – rk101 May 8 '12 at 12:48

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