Re: Non-continuous operator

Quote:

Originally Posted by

**Siron** Hi,

Let

be a Hilbert space and let

be a linear, bijective and compact operator. Prove that

is not continuous.

Hint:

is an orthonormal and total sequence in

which convergens weak to

then

as

is compact.

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

Anyone?

Thanks in advance.

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

If , what is ?

For a similar but alternative point of view, note that continuous implies bounded. Then is compact, which is possible only when is finite dimensional.

Re: Non-continuous operator

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

Since is bijective I think that .

Suppose that is continuous. Since is compact and (weak) it follows that , hence (the only element in the kernel is 0 as f is injective). Now, I have proved that weak convergence implies convergence and thus has to be finite dimensional. Which is a contradiction is if we have given that is infinite dimensional.