Operator norm with $\inf$

Question

Let $T: V \to W$ be a linear operator. The operator norm is defined as

$$ \|T\| = \sup_{v\in V: \|v\|_V = 1} \|Tv\|_W$$

Does

$$ \|T\|' = \inf_{v\in V: \|v\|_V = 1} \|Tv\|_W$$

define a norm? I believe it should and I tried to prove it but I couldn't prove that $\|T\| = 0$ implies that $T=0$.

Answer 1

Two properties of a norm are violated in general:

• we may have $\lVert T\rVert'=0$ for a non-injective $T$;
• the triangle inequality fails: if $V=W=\mathbb R^2$ and $T_1$, $T_2$ are given by the matrices $\pmatrix{1&0\\0&0}$ and $\pmatrix{0&0\\0&1}$ respectively, then $\lVert T_1\rVert'=\lVert T_2\rVert'=0$ but $\lVert T_1+T_2\rVert'=1$.

Answer 2

No, and there are lots of counterexamples. An easy one could be the following.

Let $V=\mathbb{R}^2$ equipped with any norm and $W=\mathbb{R}$. Define

$$ Tv = v_1-v_2 $$ where $v_1,v_2$ are the two components of $v$. Clearly $T$ is linear. Then if $v=[a,a]^T$, with $a$ such that $||v||_V=1$, you have $Tv=0$. Therefore

$$ \inf_{v\in V: \|v\|_V = 1} \|Tv\|_W=0 $$ but $T$ is not the null operator.