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

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.

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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.

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Thank you very much for the answer. I've been reading your solution carefully for a while, but there are things I still couldn't figure out, and hope you could kindly explain. (1) Why does the existence of a $k$ such that $\|x-Tx\| \geq k$ dist$(x,K)$ implies $\text{Im} (1-T)$ is closed? (2) Doesn't the closed graph theorem requires $F$ to be linear? (3) Which fixed point theorem are you referring to in the last sentence? How can I choose an $\bar{x}$ that gives $g(\bar{x})=0$ (not $g(\bar{x})=\bar{x}$? – DancefloorTsunderella Jan 28 '13 at 15:11
Ok then.. Which ones? :) – uforoboa Jan 28 '13 at 15:12
(As stated above. Sorry I mistakenly hit the "add comment" button before I finished writing the questions, so I edited it again.) – DancefloorTsunderella Jan 28 '13 at 15:37
As for your first point. Well... If you don't even know that it is linear there is no reasonable statement you can apply to your problem. I implicitly assumed the operator to be linear because otherwise I wouldn't know even where to start. For the second point.. yes, it is not a consequence of the fixed point theorem. I will look for the reference though – uforoboa Jan 28 '13 at 15:57
I dived into my lecture notes and found a theorem in topological degree theory that is applicable (I think it's the homotopy invariance of the Leray-Schauder degree.) Thank you very much for the solutions and clarifications! ^_^ – DancefloorTsunderella Jan 29 '13 at 12:25

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.

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Thank you very much for this counterexample, clearing the ambiguity in the problem. The statement of the problem doesn't explicitly say anything about linearity, and in class we seemed to have no agreement on assuming linearity as a default. – DancefloorTsunderella Jan 29 '13 at 12:21
I wonder if there's a counter-example that is nonlinear but continuous. I tried to come up with one in 2D (where (1) is automatically satisfied) and Brower always seemed to get in the way... – anonymous Jan 30 '13 at 9:56