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Let $E$ be a complex Hilbert space.

For $A\in\mathcal{L}(E)$, I want to show that $$\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|=\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|.$$ Clearly, since $$\{\|Ax\|;\;x\in E,\,\|x\|=1\}\subset \{\|Ax\|;\;x\in E,\,\|x\|\leq1\},$$ then $$\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\leq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|.$$

Why $$\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\geq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|?$$

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    $\begingroup$ You need to add the condition $E \ne \{0\}$, otherwise the left supremum ranges over the empty set. $\endgroup$ – gerw Jan 22 '18 at 21:59
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Hint: note that if $\|x\|\le1$, then $\|Au\|\ge\|Ax\|$ where $u=\frac{x}{\|x\|}$.

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