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!

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
$$
2 = \|Tx\|_Y \not\le \|T\|_2 \|x\|_X = 1\cdot \sqrt2.
$$
