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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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(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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