Equivalent statements of continuity of linear operators 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!
 A: (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.
