In the lecture of Inverse Problems we had the following statement:
"Let $X$ be an infinite dimensional space ($\text{dim}(X)=\infty$) and suppose that for $A\in K(X,Y)$ the inverse operator $A^{-1}$ exists, then $A^{-1}$ not continuous." (Here is $K$ the space of compact operator.)
So the things I know is (suppose now we have Banach spaces), if the map is surjective we need that the dimension of $Y$ be finite (by the open mapping theorem). But we can have that the spaces are infinite if the map is injective (e.g. embedding of $W^{1,p}$ to $L^p$ is compact). But how can we have both? Is it possible since the one space is infinite but the other not?
Let assume that they are only normed spaces (even if that it is not practicable in inverse problems, since normally we not only need Banach spaces but rather Hilbert spaces). Does it then exists?
And if it exists how we show that it is not continuous?
Thanks for the help!
Let $b$ be the closed unit ball of $X.$ Let $c=Cl_Y(A(b)).$ If $A\in K(X,Y)$ then $c$ is compact. Suppose $A^{-1}$ exists and is continuous. Then $A^{-1}(c)$ is the continuous image of $c,$ so $A^{-1}(c)$ is compact. But $b=Cl_X(b)\subset A^{-1}(c),$ so $b$ is also compact.
But $X$ has infinite vector-space dimension so $b$ cannot be compact.
There is a theorem, which I think is from Riesz, that there exists $r>0$ and an infinite $b'\subset b$ such that $\|x-y\|>r$ for all distinct $x,y\in b'.$ $(\;$If I recall correctly, $ r$ can be any member of $(0,1/4).\;)$ So $b'$ is a closed discrete sub-space of $X,$ hence $b'$ is a closed discrete sub-space of the space $b.$ But $d(x,y)=\|x-y\|$ is a metric for $b,$ and a metric space with an infinite closed discrete sub-space can't be a compact space.
Remark: A member of $K(X,Y)$ can be injective. E.g, if $X=Y=l^2$ (Hilbert space) and $A((x_n)_{n\in \Bbb N})=(x_n/n)_{n\in \Bbb N}.$