Re: Non-continuous operator

Quote:

Originally Posted by

**Siron** Hi,

Let $\displaystyle E$ be a Hilbert space and let $\displaystyle A: E \to F$ be a linear, bijective and compact operator. Prove that $\displaystyle B: F \to E, B = A^{-1}$ is not continuous.

Hint: $\displaystyle (e_n)_n$ is an orthonormal and total sequence in $\displaystyle E$ which convergens weak to $\displaystyle 0$ then $\displaystyle Ae_n \to 0$ as $\displaystyle A$ is compact.

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

Anyone?

Thanks in advance.

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

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

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

Re: Non-continuous operator

It's not explicitly given that $\displaystyle E$ is infinite dimensional. The only thing I know is that if the Hilbert space $\displaystyle E$ is infinite dimensional and separable then we can find a Hilbertbasis.

Since $\displaystyle A$ is bijective I think that $\displaystyle A^{-1}A(e_n) = e_n$.

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