1. ## Non-continuous operator

Hi,

Let $E$ be a Hilbert space and let $A: E \to F$ be a linear, bijective and compact operator. Prove that $B: F \to E, B = A^{-1}$ is not continuous.
Hint: $(e_n)_n$ is an orthonormal and total sequence in $E$ which convergens weak to $0$ then $Ae_n \to 0$ as $A$ is compact.

I don't know how I can use the hint?

Anyone?

2. ## Re: Non-continuous operator

Originally Posted by Siron
Hi,

Let $E$ be a Hilbert space and let $A: E \to F$ be a linear, bijective and compact operator. Prove that $B: F \to E, B = A^{-1}$ is not continuous.
Hint: $(e_n)_n$ is an orthonormal and total sequence in $E$ which convergens weak to $0$ then $Ae_n \to 0$ as $A$ is compact.

I don't know how I can use the hint?

Anyone?
I believe you need $E$ to be an infinite dimensional Hilbert space?

If $Ae_n \to 0$, what is $A^{-1}Ae_n$?

For a similar but alternative point of view, note that $A^{-1}$ continuous implies $A^{-1}$ bounded. Then $A^{-1}A$ is compact, which is possible only when $E$ is finite dimensional.

3. ## Re: Non-continuous operator

It's not explicitly given that $E$ is infinite dimensional. The only thing I know is that if the Hilbert space $E$ is infinite dimensional and separable then we can find a Hilbertbasis.
Since $A$ is bijective I think that $A^{-1}A(e_n) = e_n$.

Suppose that $A^{-1}$ is continuous. Since $A$ is compact and $e_n \to 0$ (weak) it follows that $Ae_n \to 0$, hence $A^{-1}A(e_n) = e_n \to A^{-1}(0) = 0$ (the only element in the kernel is 0 as f is injective). Now, I have proved that weak convergence implies convergence and thus $E$ has to be finite dimensional. Which is a contradiction is if we have given that $E$ is infinite dimensional.