inequality regarding norm of linear operator Let $T$ be a linear operator on a vector space $X$. For $x\in X$, I know there's the inequality that $$||Tx||<||T||||x||$$
Yet I'm wondering what are those norms. Are they arbitrary? especially on the right hand side there's operator norm and vector norm, how do we cooperate that?
Thank you!
 A: There are a few equivalent definitions of the operator norm. Perhaps the most relevant for this particular question is the following:
Let $T:V \to W$ be a linear map between two normed vector spaces $V$ and $W$. The operator norm $\|T\|_{op}$ of $T$ is defined to be
$$\|T\|_{op} := \sup \left\{\frac{\|Tx\|}{\|x\|} : x \in V \text{ and } x \not = 0 \right\}$$
It then follows immediately that $\|Tx\| \leq \|T\|_{op} \|x\|$ for any $x \in V$.
Check out Wikipedia for more info.
A: This inequality does not hold for arbitrary norms, but only for compatible operator and vector norms.
Let $X$ and $Y$ be normed linear spaces with norms $\|\cdot\|_X$, $\|\cdot\|_Y$, $T:X\to Y$ be linear and continuous.
Then the inequality above is true for the induced operator norm
$$
\|T\|_{L(X,Y)}:= \sup_{x\ne0} \frac{\|Tx\|_Y}{\|x\|_X}.
$$
Due to the supremum property it follows
$$
\|Tx\|_Y \le \|T\|_{L(X,Y)} \|x\|_X.
$$

Here is an example, that the inequality does not hold for arbitrary norms:
Take $X=Y=\mathbb R^2$, $\|x\|_X =\sqrt{ |x_1|^2 + |x_2|^2}$, $\|y\|_Y =  |y_1| + |y_2|$, $\|T\|=\|T\|_2$ (spectral norm):
$$
x=y = \pmatrix{1 \\1} ,  T=I_2
$$
implying
$$
2 = \|Tx\|_Y \not\le \|T\|_2 \|x\|_X = 1\cdot \sqrt2.
$$
