Continuous mapping on unit ball such that $T(\Bbb x_0)=0$

Question

got a question from a course in functional analysis.

" Let $T:\{\Bbb x\in\Bbb R: ||\Bbb x||\leq 1\}\to\Bbb R^n$ be a continuous mapping. Moreover assume that $\langle T(\Bbb x),\Bbb x\rangle>0$ for all $\Bbb x$ with $||\Bbb x||=1$. Here $\langle\cdot,\cdot\rangle$ denotes the standard inner product on $\Bbb R^n$ with the induced norm $||\cdot||$. Prove$^1$ that there exists a $\Bbb x_0\in\{\Bbb x\in\Bbb R: ||\Bbb x||\leq 1\}$ such that $T(\Bbb x_0)=0$.

$^1$Hint: Consider the mapping $G(\Bbb x)=\Bbb x - \epsilon T(\Bbb x)$ for some properly choosen $\epsilon>0$ "

From that hint it is very tempting to assume that what you should do is to show that $G(\Bbb x)$ has a fixed point in the unit ball. To do so I'd call for Brouwer's fixed point theorem, which states that any continuous function from a convex compact subset of $\Bbb R^n$ to itself has a fixed point.

If everything is correct so far, I need to show that ' $\Bbb x - \epsilon T(\Bbb x)$ ' is compact and convex. To show that it is compact, I should show that it is bounded and closed, and then, boundedness implies that it is also convex.

But now the first real problem appears (If everything is right so far); How can I show that it is bounded? In particular, that it is bounded such that $||G(\Bbb x)||\leq1$ , or that $G(\Bbb x)\in\{\Bbb x\in\Bbb R: ||\Bbb x||\leq 1\}$.

Answer 1

From that hint it is very tempting to assume that what you should do is to show that $G(\mathbb{x})$ has a fixed point in the unit ball.

Indeed. So you want to show that - for suitably chosen $\epsilon > 0$ - $G$ maps the unit ball to itself. Now, we have by assumption $\langle T(\mathbb{x}), \mathbb{x}\rangle > 0$ on the boundary of the unit ball. By compactness, there is a $c > 0$ with $\langle T(\mathbb{x}), \mathbb{x}\rangle > c$ on the boundary, and by continuity, there is a $\delta > 0$ such that $\langle T(\mathbb{x}), \mathbb{x}\rangle > c$ for all $\mathbb{x}$ with $\lVert \mathbb{x}\rVert \geqslant 1 - \delta$.

Now find an $\epsilon > 0$ such that

$$\lVert G(\mathbb{x})\rVert^2 = \lVert \mathbb{x}\rVert^2 - 2\epsilon\langle T(\mathbb{x}),\mathbb{x}\rangle + \epsilon^2 \lVert T(\mathbb{x})\rVert^2 \leqslant 1$$

for all $\mathbb{x}$ in the closed unit ball.

Answer 2

Thank you very much for your answer, Daniel!

So now for a final answer, can I simply replace every element describing $\lVert G(\mathbb{x})\rVert^2$ by those numbers ($\delta$ and $c$) and using normal calculus to solve for $\epsilon$ ?

That is we had...

$\lVert G(\mathbb{x})\rVert^2 = \lVert \mathbb{x}\rVert^2 - 2\epsilon\langle T(\mathbb{x}),\mathbb{x}\rangle + \epsilon^2 \lVert T(\mathbb{x})\rVert^2 \leqslant 1$

...and then $\lVert \mathbb{x}\rVert^2$ is replaced by $(1-\delta)^2$, $\langle T(\mathbb{x}),\mathbb{x}\rangle$ by $c$ and, since $T(\Bbb x)$ is continuous, $\|T(\Bbb x)\|^2$ can be replaced by it's maximum on the unit ball which is finite; $sup_{\|\Bbb x\|\leq1}\{\|T(\Bbb x)\|^2\} = M< \infty$

If so, then by simple rearrangements, $\epsilon\in [0,\frac{c+\sqrt(c^2+M(1-(1-\delta)^2))}{M}]$

Seems legit?

Also, are you supposed to "tag" the one you're replying to? @Daniel Fischer