# Inequality related to the norm of operator

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 $$a \leq \|T\|$$?

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 $$a \leq \vert\vert T \vert\vert$$.
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 $$\left\lVert \frac{x}{\left\lVert x\right\rVert}\right\rVert=1.$$