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I am asked to prove that the following are true:

Given a linear operator $T: X \to Y$ where $X,Y$ normed linear spaces:

(1) $T$ continuous at at point $\iff$ $T$ continuous everywhere

(2) $T$ continuous $\iff$ $T$ Lipschitz.

(3) Show that the above fail if $T$ is not linear.

I have (1) entirely and I have the reverse of (2). I need some help with the forward of (2) and also some examples of non-linear operators because I know none.

Thank you very much in advance!

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Regarding (3), can you think of any function, say $\mathbb R\to\mathbb R$, that is continuous at a point but not continuous everywhere? Can you think of any function, say $\mathbb R\to\mathbb R$, that is continuous but not Lipschitz? How about continuous but not uniformly continuous? –  Jonas Meyer Feb 21 '13 at 6:08
    
Ah, so I am free to chose the normed spaces, too, when coming up with counterexamples? I though I had to restrict myself to arbitrary normed linear spaces! Thanks for the hint! –  user44069 Feb 21 '13 at 6:10
    
You're welcome. I don't know, it is a little ambiguous about whether you're supposed to show that there are counterexamples regardless of the spaces (assuming they're not $0$-dimensional), but either way it is good to start with something simple and concrete I think. –  Jonas Meyer Feb 21 '13 at 6:14

1 Answer 1

(3) Let $B$ be the open unit ball centered at $0$ in $X$. And fix a point $y\neq 0$ in $Y$. Then the function $$ T(x)=1_B(x)y $$ is constant on $B$, hence continuous on $B$. But it is clearly discontinuous.

So (1) fails if $T$ is not linear.

Now consider the function $$ T(x)=e^{\|x\|}y. $$ Since the norm is continuous, this is continuous by composition of continuous functions. Now if it was C-Lipschitz, we would get $$ \|T(x)-T(0)\|=(e^{\|x\|}-1)\|y\|\leq C \|x\| $$ for all $x\in X$. The contradiction follows when letting $\|x\|$ tend to $+\infty$.

So (2) fails if $T$ is not linear.

(2) You did the forward implication. So now assume $T$ is continuous. Recall that $B$ denotes the open unit ball centered at $0$.

The set $T^{-1}(B)$ is open and contains $0$. So it contains an open ball centered at $0$. Such a ball is of the form $kB$ by homogeneity of the norm.

So $$ kB\subseteq T^{-1}(B). $$ This means that $\|x\|<k$ implies $\|Tx\|<1$.

Now for an arbitrary nozero $x$, $$ k\frac{x}{\|x\|}\in kB \quad\mbox{hence}\quad \left\|T\left( k\frac{x}{\|x\|} \right)\right\| <1. $$ It follows that $$ \|T(x)\|\leq \frac{1}{k}\|x\| $$ for all $x\in X$.

Since $T$ is linear, this yields obsviously that $T$ is Lipschitz.

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