Skip to main content

Existence through Fixed Points

06 Sep 2020 - Tags: quick-analysis-tricks

While watching this lecture (in case the link breaks one day: Steffano Luzzatto teaching Dynamical Systems at ICTP, lecture 11), I saw my first clever use of fixed-point theorems in analysis. In particular, we used a leveled-up version of the contraction mapping theorem to show that a certain map was actually a homeomorphism. The main takeaway is that by rearranging what you want to be true into a statement that some object is a fixed point, we can show that our dreams can be realized. This mode of argument is extremely common in algebra and logic, so it’s like a friendly face in the often stressful world of analysis.

We start with the leveled-up contraction mapping theorem. This theorem lets us take a family of contractions (parametrized by some metric space) and conclude that the fixed points vary continuously in the parameter.

Let $\Omega$ and $X$ be metric spaces, with $X$ complete. Let $f : \Omega \times X \to X$ be continuous so that each $f(\omega,-)$ is contracting with the same constant $\lambda$. Then the fixed point $p_\omega$ of $f(\omega,-)$ varies continuously in $\omega$.

In the interest of completeness (pun intended), the proof goes as follows (this is basically taken straight from the linked video):

We will write $f_\omega(x)$ for $f(\omega,x)$. For any $x \in X$, we have $|x - p_\omega| \leq |x - f_\omega(x)| + |f_\omega(x) - p_\omega|$. Then since $p_\omega = f_\omega(p_\omega)$, we see this is the same as $|x - p_\omega| \leq |x - f_\omega(x)| + |f_\omega(x) - f_\omega(p_\omega)|$. Lastly, since $f_\omega$ is a contraction, $|x - p_\omega| \leq |x - f_\omega(x)| + \lambda|x - p_\omega|$, which we rearrange to $|x - p_\omega| \leq \frac{|x - f_\omega(x)|}{1-\lambda}$.

Then by specializing to $x = p_{\eta}$, we see $|p_{\eta} - p_\omega| \leq \frac{|p_{\eta}-f_\omega(p_{\eta})|}{1-\lambda}$. But as $\eta \to \omega$, the right hand side goes to $0$ (by the continuity of $f$). So $|p_{\eta} - p_\omega|$ goes to $0$ too, showing $\omega \mapsto p_\omega$ is continuous.

Notice that the above proof uses the notation of a Banach Space, but never uses the linear algebraic structure. Even though I wrote $|x-y|$, the proof only uses the fact that $|x-y|$ is a complete metric, and you could write $d(x,y)$ throughout instead. This is a proof for arbitrary (complete) metric spaces, despite what the notation might lead you to believe.

Now for the surprising application! We want to show that if you perturb an invertible linear transformation, it remains a homeomorphism. Again, the clever idea that I want to highlight in this post is this: by recognizing a hypothetical inverse of a point $y$ would have to be a fixed point of some function $f_y$, then we win as soon as we show $f_y$ is a contraction. Moreover, since these $f_y$ will depend continuously on $y$, the previous theorem tells us that our inverse will depend continuously on $y$!

So here’s the actual theorem proven in the lecture:

Let $E$ be a Banach Space, and let $T : E \to E$ be an invertible linear contraction. Let $e : E \to E$ be lipschitz with constant $\lambda < ||T^{-1}||^{-1}$. Then $T + e$ is still a homeomorphism.

The condition on $T$ makes sense. We clearly want to use the contraction mapping theorem, so if $T$ is supposed to be the “main term” of this function, it had better be a contraction. It’s not immediately clear why this condition on $e$ means that it is a “small” perturbation. However, if we unwind the definition of the operator norm, this condition says that $e$ has lipschitz constant less than $\inf_{||v||=1} ||Tv||$. So if I wiggle $x$ a little, $ex$ is wiggled by less than $\inf ||Tv||$. In particular, $ex$ is wiggled by less than $||Tx||$. So the amount of extra wiggle incurred by $e$ is small compared to the wiggle that $T$ already has.

Primed by the idea that we’ll be exploiting the contraction mapping theorem, the proof becomes almost obvious. We want to find a continuous inverse for $T+e$, but what must such an inverse look like?

\[\begin{align*} (T+e)x = y &\iff Tx + ex = y\\ &\iff Tx = y-ex\\ &\iff x = T^{-1}y - T^{-1}ex \end{align*}\]

So a hypothetical inverse $x$ must be a fixed point of $f_y(x) = T^{-1}y - T^{-1}ex$. Notice $f_y(x)$ is jointly continuous in $y$ and $x$, since subtraction is. So by our previous theorem, it suffices to show each $f_y$ is a contraction, and that there is a uniform contraction constant. Of course (independtly of $y$)

\[\begin{align*} ||f_y(x) - f_y(x')|| &= ||T^{-1}ex' - T^{-1}ex||\\ &= ||T^{-1}(ex' - ex)||\\ &\leq ||T^{-1}|| \; ||ex' - ex||\\ &\leq ||T^{-1}|| \; \lambda \; ||x' - x||\\ &\leq ||x' - x|| \end{align*}\]

Here the last inequality came from our (previously mysterious) condition on the lipschitz constant of $e$. Putting ourselves in the shoes of someone trying to prove this in their research, it is now clear how that condition arose in the first place. We might try to work through this proof, and in the last step realize that we need this condition on $e$. So we add it as a hypothesis, and pretend we knew in advance that this was the right condition!