Under what conditions we have $\|Tx\| = \|T\| \|x\|$? For bounded linear operator we have  $\|Tx\| \le \|T\| \|x\|$ but when the equality takes, i.e., $\|Tx\| = \|T\| \|x\|$? 
PS. 
If $T$ is an isometry then $\|Tx\|=\|x\|$ and $\|T\|=1$, therefore the equality holds for all $x$, right? 
 A: This problem exhibits a different behavior in the Hilbert case than in the general case. 
Let $T\colon X\to Y$ be a bounded linear operator. The problem is to understand whether the set 
$$
M:=\{ x\in X\ :\ \|Tx\|=\|T\|\|x\|\}
$$ 
is reduced to $\{0\}$ or not. This set is sometimes known as the set of norming vectors, or the set of maximizers of $T$.
Now, if $X, Y$ are Hilbert, the problem is reduced to the spectral analysis of the nonnegative definite operator $T^\star T$. Indeed, the intersection of $M$ with the unit sphere $S\subset X$ is 
$$
M\cap S =\{ x\in X \ :\ \|Tx\|^2\ \text{is maximal, given that }\|x\|^2=1\},$$ 
and by the Lagrange multiplier theorem, its elements must satisfy 
$$\tag{1}
d(\|T\cdot\|^2)(h)=\mu d(\|\cdot\|^2)(h),\, \qquad \forall h\in X, $$
where $d$ denotes the Fréchet derivative. Since we are in a Hilbert setting, we can compute such derivatives using that $\|Tx\|^2= \langle T^\star Tx|x\rangle$ and $\|x\|^2=\langle x|x\rangle$. Equation (1) thus reads 
$$
\tag{2} 
\langle T^\star T x| h\rangle = \mu \langle x|h\rangle,\qquad \forall h\in X, $$ 
which is the eigenvalue equation
$$
T^\star T x = \mu x.$$
Taking $h=x$, we also see that $\mu=\|Tx\|^2=\|T\|^2.$ So $\mu$ must be maximally large.

We conclude that $M$ is the set of eigenvectors of $T^\star T$ corresponding to the largest eigenvalue. It can be that $T^\star T$ has no eigenvalues, in which case $M=\{0\}$. 


In the Banach space case, the problem is more complicated. The set $M$ needs not be an eigenspace, it needs not even be a vector space. For example, if $T\colon H^1(\mathbb R)\to L^\infty (\mathbb R)$ is the Sobolev embedding of 
$$H^1(\mathbb R)=\{ f\in L^2(\mathbb R)\ :  f'\in L^2(\mathbb R)\}, $$ 
then 
$$
M=\{ A\exp(-|\cdot +x_0|)\ :\ A\in \mathbb R\, x_0\in\mathbb R\}, $$ 
and this is not a vector space. Similar examples can be obtained by considering other Sobolev embeddings.
