real analysis - Continuous functional on the linear operator

Let $\Pi, \hat \Pi$ be two linear operators from $U$ to $V$. The norm-distance is defined as

$$||\hat \Pi- \Pi||=\sup_{x\in U}\frac{||(\hat \Pi- \Pi)x||}{||x||}$$

Let us define a continuous bounded function $g:V\rightarrow \mathbb{R}$,

Claim: So for every $\epsilon>0$, $\exists \delta>0$ such that

$$\sup_{x\in U}\frac{||(\hat \Pi- \Pi)x||}{||x||}<\epsilon\Rightarrow \sup_{x\in U}\frac{||g(\hat \Pi x)- g(\Pi x)||}{||x||}<\delta.$$

Is the claim is right based on the "continuity" and bounded (only) assumption of $g$? Or do we need uniform continuity as well?

Thanks.

Answer

The claim is not true for merely continuous and bounded $g$. For instance, take

$$g(v):=\min( \sqrt{\|v\|},1).$$
Fix $x$ such that $\Pi x \ne \hat\Pi x$.
Then for $s>0$ such that $\|s\hat\Pi x\|\le1$ and $\|s\Pi x\|\le1$
$$\|g(s\hat\Pi x)-g(s\Pi x) \|= \sqrt s\left|\sqrt{\|\hat\Pi x\|} - \sqrt{\Pi x}\right|,$$
hence for $s\searrow 0$ the quantity

$$\frac{\|g(s\hat\Pi x)-g(s\Pi x) \|}{\|sx\|}$$
blows up.

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If in addition, $g$ is globally Lipschitz continuous, i.e., $|g(v_1)-g(v_2)|\le L \|v_1-v_2\|$ for all $v_1,v_2\in V$ the claim can be proven easily.