How to prove $\|Tx\|<\|T\|$ where T is bounded linear operator

Question

I have bounded linear operator $T$ and an element $x$ belong to domain of $T$. Now I am going to prove that there is strict inequality $\left \|Tx \right \|< \left \|T \right \|$ As we know that for any operator $T$ we can write $\left \|Tx \right \| \leq \left \|T \right\|\cdot \left \|x \right \| \leq \left \|T \right \|$. My question how to prove strict inequality

Answer 1

This is false. For example, consider $T=1_X$ where $X$ is a normed space. Then $\Vert T \Vert = 1$ yet $\Vert Tx \Vert = \Vert x \Vert$ can be made arbitrarily large.

Answer 2

If $T$ is an isometric isomorphism then $\|Tx\|=\|x\|=\|T\|\|x\|$ for all $x$

If $\|x\|<1$ and $T$ is any non-zero operator then $\|Tx\|\leq \|T\|\|x\|<\|T\|$ so strict inequality is true.

Answer 3

It depends on the question asked : whether there is no restriction imposed on the $\Vert x \Vert$ - the statement is false, because you may rescale $x$ by an arbitrary factor. Whether it is assumed, that $\Vert x \Vert = 1$, then, by definition of operator norm: $$ \Vert T \Vert = \sup_{\Vert x \Vert = 1} \Vert T x \Vert $$ we have non-strict inequality in general. In case the value on which $\sup$ is reach, doesn't belong to the domain, the inequality becomes strict.