Given a linear function $A$ between two normed Vectorspaces i have to show euquality of the follwing statements:
- $A$ is continuous
- There exists a point where $A$ is continuous
- $A$ is Lipschitz-continuous
$3\to1\to2 \:$ is obviously true, but i can't find any other relation.
Thanks in advance
3 $\Rightarrow$ 2 Let $\varepsilon >0$. Let $L$ be such that $\forall x,y ||A(x-y)|| \le L \cdot||x-y||$. In particular, if you take $y=0$, $\forall x ||A(x)|| \le L ||x||$. Then, if $\eta = \varepsilon / L$, $||x|| < \eta \Rightarrow ||A(x)|| \leq \varepsilon$, so that $A$ is continuous in $0$.
2 $\Rightarrow$ 1 If $A$ is continuous at $x_{0}$, then $A$ is continuous at any point $x$ because $$A(x+h) - A(x) = A(x+h-x) = A(h) = A(x_{0} + h - x_{0}) = A(x_{0} + h) - A(x_{0})$$ which tends to $0$ as $h \rightarrow 0$ because of the continuity at $x_{0}$.
1 $\Rightarrow$ 3 $A$ is continuous at $0$, so that $\exists \eta, h < \eta \Rightarrow ||A(h)|| = ||A(h) - A(0)|| \leq 1$. Now for any $x, y \in X$, $$||A(x) - A(y)|| = ||A(x-y)|| = || \dfrac{||x-y||}{\eta} A\left(\dfrac{(x - y)}{||x-y||/\eta}\right)|| \leq \dfrac{||x-y||}{\eta}.$$