I have a question about functional analysis. I am currently working on the following two questions.
In the following, K(X,Y) denotes the set of all compact operators from X to Y whose domain of definition is the whole of X. B(X,Y) denotes the set of all bounded linear operators from X to Y whose domain of definition is the whole of X. R(K) denotes the image of the operator on K.
Questions
[1] Suppose $X,Y$ are Banach spaces and $K¥in K(X,Y)$. If $R(K)=Y$, show that $Y$ is finite dimensional.
[2] For $X,Y$ Banach spaces, let A be an operator in $B(X,Y)$ such that $R(A)$ is closed and infinite dimensional. Show that A is not compact.
My view
In [1], we are going to show that the unit sphere $S$ of $Y$ is compact and that $Y$ is finite dimensional.
In [2], assuming that $A$ is compact, I am going to show that the range of $A$ is finite dimensional, i.e., the unit sphere $S$ of $A$ is compact, and derive a contradiction.
In both cases, any point sequence {$y_n$} of $S$ can be written as $y_n=Kx_n$, so that $||Ax_n||=1$.
This is where I now have a problem. If I can show that such {$x_n$} is a bounded sequence of points in $X$, I can solve both problems, but I just can't show this.
I think the reason why such {$x_n$} can be solved if it is a bounded point sequence is as follows. It is enough to show that $Y$ is a finite dimensional space, and that the unit sphere $S$ of $Y$ is a compact set. Take a sequence of points {$y_n$} in $S$. From $R(K)=Y$, for each $n=1,2,...$. For each $n=1,2,...$, we get {$x_n$}$¥subset X$ with $y_n=Kx_n$. If this {$x_n$} is a bounded sequence of points in $X$, then by the compactness of the operator $K$, {$Kx_n$} has a subsequence that converges to a point in $Y$. In other words, {$y_n$} has a subsequence that converges to a point in $Y$. Here, for each $y_n$, since $||y_n||=1$, the convergent subsequence $y$ converges to $||y||=1$. Therefore, $y∈S$. This shows that $S$ is a compact set.
Is there something fundamentally wrong with the policy? I apologize for the length of this article, and if anyone can help me, I would appreciate it.
For 1): $K$ is an open map by Open Mapping Theorem. So $K(\{x:\|x\|<1\})$ contains an open ball around $0$, say, $\{y: \|y\|<r\}$ in $Y$. But $K$ is compact so this ball is relatively compact which means $\{y: \|y\|\leq r \}$ is compact. Hence, $Y$ is finite dimensional.