Given $F:X \to X$ on $X$:Hilbert space satisfying some properties, prove that $F$ is surjective.

Question

Given a map $F:X \to X$ where $X$ is a Hilbert space, $F$ satisfying

1. $f(x):=x-F(x)$ is a compact map.
2. $\lim_{\|x\|\to \infty} \frac{(F(x),x)}{\|x\|} = \infty$

I'm seeking to prove that $F$ is surjective, i.e.

$$\forall y \in X,\exists x \in X (F(x)=y)$$

Any hints or suggestions? Thank you in advance.

PS. To prove the existence of a solution to $F(x)=y$, I'm thinking of using some form of the fixed-point theorem, so I'm wondering if it can proven that $f$ (and $F$) is continuous. Also, I'm at a loss at 2. What use is the limit? Anyway, I have completely no idea how to solve this. Please help.

Answer 1

Lemma: Assume that $X$ is a Banach space, and let $T:X\to X$ be compact. Then $\text{Im}(1-T)(X)$ is closed in $X$.

Proof: Set $K:=\ker(1-T).$ We will show that there exists $k>0$ such that $\|x-Tx\|\geq k\;\text{dist}(x,K)$, for every $x\in X$.

Assume such a $k$ does not exist.Then there is a sequence ${x_n}\in X\setminus K$ such that $$\lim_{n\to\infty}\frac{\|x_n-T x_n\|}{\text{dist}(x_n,K)}=0.$$ As $K$ is finite dimensional (because $T$ induces the identity on it, and the identity is compact only if the space is finite dimensional), its closed bounded subsets are compact (Heine Borel), and therefore for every $n\in\mathbb N$ there exists $z_n$ such that $$\|x_n-z_n\|=\text{dist}(x_n,K).$$ Since $z_n\in K,$ we can also write $$x_n-T x_n=x_n-z_n-T(x_n-z_n),$$ so that $$\frac{\|x_n-T x_n\|}{\text{dist}(x_n,K)}=\frac{\|x_n-z_n-T(x_n-z_n)\|}{\|x_n-z_n\|}=\left\|\frac{x_n-z_n}{\|x_n-z_n\|}-T\left(\frac{x_n-z_n}{\|x_n-z_n\|}\right)\right\|=\|u_n-T(u_n)\|.$$ Notice that $\|u_n\|=1$ and $\text{dist}(u_n,K)=1$. But as $T$ is compact, $T u_n$ has a converging subsequence, say to $v$. Since $\lim_{n\to\infty} u_n-T(u_n)=0,$ it follows that also $u_n\to v$. Then $v\in K$ and this is absurd since $\text{dist}(u_n,K)=1$.

Now, $f$ is compact, then by the lemma $F(X)$ is closed in $X$. By the closed graph theorem, $F$ is continuous.

Choose then any $y\in X$ and consider $g(x)=F(x)-y$. Choose a ball centered at $0$ so that it is an open neighborhood of $y$ and moreover $g(x)$ is not $0$ on the boundary of this ball, and this is exactly where your second hypothesis comes into play (you can take $R$ very big so that all these things are true).

Finally use the topological degree theory to conclude that there exists $\bar x\in X$ such that $g(\bar x)=0$ i.e. $F$ is surjective.

Answer 2

If you do not assume $F$ to be linear or continuous, then it will in general not be surjective: Choose $X = \mathbb R$ and $$F(x) = \begin{cases} x & \text{if $|x| \ge 1$,}\\ 0 & \text{otherwise.} \end{cases} $$ Now $$f(x) = \begin{cases} 0 & \text{if $|x| \ge 1$,}\\ x & \text{otherwise,} \end{cases} $$ so that $f$ is compact (since the image of any ball is contained in $[-1,1]$). We also have $F(x)x = x^2$ whenever $|x| \ge 1$; in summary, both conditions are fulfilled but $F$ is not surjective.