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.
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.