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Given a linear function $A$ between two normed Vectorspaces i have to show euquality of the follwing statements:

  1. $A$ is continuous
  2. There exists a point where $A$ is continuous
  3. $A$ is Lipschitz-continuous

$3\to1\to2 \:$ is obviously true, but i can't find any other relation.

Thanks in advance

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  • $\begingroup$ I got the tipp to write down the smaler side of the Lipschitz-equation and use linearity but i dont get anywhere (Let $X$ and $Y$ be the normed spaces with $A: X \to Y$$$\textrm{$A$ is Lipschitz}\:\: \Leftrightarrow \:\:\forall _{x,y\in X}\: \: ||A(x)-A(y)||_Y\le L\cdot||x-y||_X $$ $$\Rightarrow ||A(x-y)||_Y \le L \cdot||x-y||_X$$ what can i gain from this? $\endgroup$ – DeltaChief May 28 '16 at 20:44
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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 $x, y \in X$, $$||A(x) - A(y)|| = ||A(x-y)|| = || \dfrac{||x-y||}{\eta} A\left(\dfrac{(x - y)}{||x-y||/\eta}\right)|| \leq \dfrac{||x-y||}{\eta}.$$

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