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

  • 1
    $\begingroup$ $T:\{0,1\}\to \{0,1\}$ defined by $T(0)=1$ and $T(1)=0$ does not have fixed points. $\endgroup$
    – Jochen
    Apr 11, 2019 at 8:22

1 Answer 1


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.

  • $\begingroup$ Rotation matrix, right? $\endgroup$ Apr 10, 2019 at 17:54
  • $\begingroup$ 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! $\endgroup$ Apr 10, 2019 at 17:56

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