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The question is simple: Suppose $f : \mathbb{R}^ 2 \to \mathbb{R}^ 2$ is continuous. Show that there exist $\lambda > 0$ and $x \in \mathbb{R}^2$ such that $f(x) = \lambda x$.

So basically, we have to show that there is at least one point that gets scaled by $f$. If $f$ has a fixed point $x^*$, then we are done, since we can then take $x = x^ *$ and $\lambda = 1$ and we are done. What we could do is arrive at a contradiction when assuming that the following is not true: for all $\lambda > 0$, we have that $(1/\lambda) f(x)$ has no fixed point $x^*$. Because then, by contradiction, there is a $\lambda$ such that the function $(1/\lambda) f(x)$ has a fixed point $x^*$, so that $f$ had to scale $x^*$ by exactly $\lambda$, which is what we wanted.

However, why should we arrive at contradiction? Since $f$ has no fixed point, why is it a problem if $(1/\lambda) f$ doesn't have one either?

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  • $\begingroup$ Do you know any fixed-point theorems? Have you tried adapting your situation to fit the conditions of such theorems? $\endgroup$ – Mees de Vries May 10 '16 at 14:19
  • $\begingroup$ Yes, I know about the fixed-point theorems by Banach and Brouwer. Is there another useful one? $\endgroup$ – limitIntegral314 May 10 '16 at 14:20
  • $\begingroup$ The one by Brouwer ought to do it. (Of course, you have to fit your situation into the conditions of the theorem.) Hint: you can even prove that $x$ lies in the unit disc. $\endgroup$ – Mees de Vries May 10 '16 at 14:24
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    $\begingroup$ I think I got it. If $g$ is the restriction of $f$ to the unit disk, look at the image of $g$ and then divide all elements in said image by $\lambda$, the length of the vector that is the furthest away from the origin. Then $g$ divided by $\lambda$ is a function from the unit disk to the unit disk, and hence has a fixed point, which means $g$ (and hence $f$) had to scale that fixed point by $\lambda$. $\endgroup$ – limitIntegral314 May 10 '16 at 14:30

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