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I have $H$ an Hilbert Space and $L:(H,\left\| \cdot \right\|_1) \rightarrow (H,\left\| \cdot \right\|_2)$ linear and bijective; here $\left<x,y\right>_2:=\left<Lx,Ly\right>_1$ and so $\left \| x \right \|_2:=\left \| Lx \right \|_1$

Where can I find the proof of the following? $$\left\| \cdot \right\|_1 \mbox{is equivalent to} \left\| \cdot \right\|_2 \leftrightarrow L \mbox{ is continuous}$$

Skills
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  • What does $\langle Lx, Ly \rangle_1$ mean, actually? How is $\langle .,.\rangle_1$ even defined? – 5xum Oct 22 '15 at 17:48
  • maybe now is better. – Skills Oct 22 '15 at 17:52
  • No, you still defined $\langle x,y\rangle_2$ using $\langle .,.\rangle_1$, without telling us what $\langle .,.\rangle_1$ actually is. – 5xum Oct 22 '15 at 17:54
  • an inner product defined in $H$.. I don't know what tell you – Skills Oct 22 '15 at 17:58
  • But if $L$ is defined, and $\langle .,.\rangle_1$ and $\langle .,.\rangle_2$ is defined, then what exactly did you define when you said $$\langle x,y\rangle_2=\langle Lx, Ly\rangle_1?$$ – 5xum Oct 22 '15 at 18:00
  • But the second one is defined using the first one – Skills Oct 22 '15 at 18:04

1 Answers1

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If $\|.\|_1$ is equivalent to $\|.\|_2$ then there is a constant $C$ with $\|x\|_2 \leq C \|x\|_1$ for all $x \in H$ so we have $$\|Lx\|_2 \leq C \|Lx\|_1 = C \|x\|_2 \leq C^2 \|x\|_1$$ for all $x \in H$ which implies that $L$ is continuous.

If $L$ is continuous then $\|x\|_2 = \|Lx\|_1 \leq \|L\| \|x\|_1$ and $\|x\|_1 = \|L^{-1} x\|_2 \leq \|L^{-1}\| \|x\|_2$. So the norms are equivalent.

user282912
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  • where $\left | L \right |$ is the sup of $Lx$? – Skills Oct 22 '15 at 19:47
  • Why is $L^{-1}$ continuous if $L$ is continuous and bijective? I see it in finite dimensions, but why does this work in infinite dimensions? – Ian Oct 23 '15 at 00:00