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Let $T$ a bounded operator acting on a complex Hilbert space $\mathcal{H}$.

It is well know that the operator norm of $T$ is given by:

$$\|T\|=\sup\left\{\frac{\|Tx\|}{\|x\|}\,;\;x\in \mathcal{H},\,x\neq 0\right\}.$$

Let $$ a=\sup A,\;\text{with}\;\;\; A=\left\{\|Tx\|\,;\;x\in \mathcal{H},\,\|x\|\leq 1\right\}.\\ $$

I want to prove that $$a \leq \|T\|.$$

Let $x\in \mathcal{H}\setminus \{0\}$ be such that $\|x\|\leq 1$, then we have $$\|Tx\| \leq \frac{\|Tx\|}{\|x\|}.$$ So, $$\|Tx\| \leq \|T\|,$$ for all $x\in \mathcal{H}\setminus \{0\}$ with $\|x\|\leq 1$.

How to prove that $a \leq \|T\|$?

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You are done basically. What you have shown is that $$\forall b \in A: b \leq \vert\vert T \vert\vert$$ As the inequality holds for every $b \in A$, it also holds for the supremum of $A$, which is denoted as $a$ in your notation. So you have $a \leq \vert\vert T \vert\vert$.

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In fact, these are the same quantity. If $x\neq 0$ and $\left\lVert x\right\rVert\leq 1,$ then $\left\lVert Tx\right\lVert\leq \frac{\left\lVert Tx\right\rVert}{\left\lVert x\right\rVert}\leq \left\lVert T\right\rVert $, and so $a\leq \left\lVert T \right\rVert\,$ since the above is true for any $x\neq 0$ with $\left\lVert x\right\rVert\leq 1.$ Now, for the other inequality:

Note that for any $x\neq 0,$ $$\frac{\left\lVert Tx\right\rVert}{\left\lVert x\right\rVert}=\left\lVert T\left(\frac{x}{\left\lVert x\right\rVert}\right)\right\rVert\leq a,$$ since $\left\lVert \frac{x}{\left\lVert x\right\rVert}\right\rVert=1.$

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