Brouwer fixed point theorem : continuous case

Let $\overline{P} = \frac{P}{\delta/2+1}$. $\overline{P}$ takes its values in $D^n$ as for $x \in D^n$: $\left\Vert P(x) \right\Vert \le \delta/2+1$.

You have for $x \in D^n$

$$\begin{aligned} \left\Vert \overline{P}(x) - G(x) \right\Vert &= \left\Vert \frac{P(x)}{\delta/2+1} - G(x) \right\Vert\\ &=\frac{1}{\delta/2+1}\left\Vert (P(x) - G(x)) - (\delta/2) G(x) \right\Vert\\ &\le \frac{1}{\delta/2+1}\left(\left\Vert (P(x) - G(x))\right\Vert + (\delta/2) \left\Vert G(x) \right\Vert\right)\\ &\lt \frac{1}{\delta/2+1}\left(\delta/2+ \delta/2 \right)\lt \delta \end{aligned}$$

And if $\overline{P}$ was having a fixed point $x_0$, you'll get the contradiction

$$ \left\Vert x_0 - G(x_0) \right\Vert=\left\Vert \overline{P}(x_0) - G(x_0) \right\Vert \lt \delta$$