# Convexity is crucial in Schauder-Tychnonov fixed-point theorem.

The following Theorem is well-known:

Schauder-Tychnonov fixed-point theorem: Let $$K$$ be a compact convex subset of a Banach space, $$E$$. If $$T:K\to K$$ is continuous, then $$T$$ has a fixed point.

I'm trying to see that if I relax convexity, then the Theorem does not hold. I'm thinking of a sphere in the space, $$\left(C[0,1],\|\|\right)$$ as counter-example.

However, I don't know how to get it together. Any help in putting it together, will be appreciated. Thanks

• $T:\{0,1\}\to \{0,1\}$ defined by $T(0)=1$ and $T(1)=0$ does not have fixed points. – Jochen Apr 11 at 8:22

## 1 Answer

Consider the map: $$T: S^1 \to S^1$$ given by quarter-clockwise rotation (here $$E$$ is $$\mathbb{R}^2$$), where $$S^1 = \{x \in \mathbb{R}^2 : \|x\| =1\}$$.

This is a fixed-point free, continuous map from a compact (albeit not convex) subset to itself.

Edit: What really is needed is some sort of 'no holes' type property for your domain. Convexity is one condition that ensures this and as a condition gives you a lot of fine analytic structure to work with, but in the spirit of the question, it is 'far' from necessary.

• Rotation matrix, right? – Omojola Micheal Apr 10 at 17:54
• Just define it in polar coordinates. Our domain is $\{(r, \theta) : r = 1\}$. Then $T(1,\theta) = (1, \theta + \frac{\pi}{2})$. You can also think of it as a rotation restricted to the circle though! – Pete Caradonna Apr 10 at 17:56