So this question had two main parts that I got stuck on:

Suppose that (X,d) is a complete metric space and $f : X \rightarrow X$ is a map.

Parts a) & b) just asked for the definition of a contraction and to prove that $f$ has at most one fixed point without using Banach's fixed point theorem, which I was fine with.

(c) Prove that $f : \mathbb{R} \rightarrow \mathbb{R}, x\mapsto f(x)= $ $\frac{1}{20} \frac{1}{1+x^4}$ is a contraction.

(d) Use the Banach fixed point theorem to prove that the polynomial equation $x^5 + 3x − 1 = 0$ has exactly one real solution and compute this solution numerically to 3 decimal places.

So for part c) I have:

For $C^1$ functions $|f(x)-f(y)|\leqslant M|x-y|$ if $|f'(x)|\leqslant M$.

We compute

$$f'(x)= -\frac{x^3}{5(x^4+1)^2}$$

$$=-\frac{x^3}{(x^4+1)^2}\cdot\frac{1}{5}$$ $$\leqslant \frac{1}{5}$$

Therefore $$|f(x)-f(y)|\leqslant\frac{1}{5}|x-y|$$and hence $f$ is a contraction.

If somebody could tell me if this is correct I would appreciate it a lot!

Part d) I am completely stuck on and don't really know how to tackle it! All I managed to do was compute the root to be 0.332 by iterating.

  • $\begingroup$ You need to show $|f’(x)| \leq 0.2$, not just $f’(x) \leq 0.2$, and certainly not under the assumption that “$|f’(x)| \leq M$”. $\endgroup$ – Mindlack Aug 21 at 20:50
  • $\begingroup$ @Mindlack The conditional “if $|f'(x)| ≤ M$” belongs to the statement in the line above. It should be followed by a period like this “… if $|f'(x)| ≤ M$. We compute …” $\endgroup$ – k.stm Aug 21 at 20:52
  • $\begingroup$ @k.stm Ah yes sorry my bad, I'll try and fix it. $\endgroup$ – hiling99 Aug 21 at 20:55
  • $\begingroup$ @Mindlack Thanks, I'll have a go at doing that. $\endgroup$ – hiling99 Aug 21 at 21:00
  • $\begingroup$ My edit was for typos in the statement of c). BTW I prefer to type f:\Bbb R\to \Bbb R which gives $f:\Bbb R\to \Bbb R$, and x\to f(x), which gives $x\to f(x)$ with a lot less typing. Also, when \mathbb or \Bbb or most things like them are applied to a single keystroke like R, you do not need {} braces. Just put a space before the R. BTW a hard marker might want you to prove that $|x^3/(5(1+x^4)^2)|\le 1/5$ for all $x\in \Bbb R.$ $\endgroup$ – DanielWainfleet Aug 22 at 0:47

Yes the first part is (mostly) fine, you have by the mean value theorem $$d(f(x),f(y)) = |f(x) - f(y)| \leq \sup_{z\in\mathbb{R}}|f'(z)||x-y|\leq \frac{1}{5}|x-y| = \frac{1}{5}d(x,y)$$ Therefore it is a contraction. For the next part, consider

$$g(x) := \frac{1}{3+x^4}$$ If $g$ has a unique fixed point $x_0$ (by the Banach fixed point theorem), then $$g(x_0) = x_0 = \frac{1}{3+x_0^4} \iff x_0^5 + 3x_0 - 1 =0$$ as in the proof of the Banach fixed point theorem, you may take any point, say $x=0$, then $$x_0 = \lim_{n\rightarrow\infty}g^n(x)$$ Where I am using the notation $g^{n}(x) = g^{n-1}(g(x))$.

  • $\begingroup$ Thanks! This might seem a really stupid question but how do you get g(x) in the first place? $\endgroup$ – hiling99 Aug 21 at 21:19
  • $\begingroup$ You can just define it like that and re-compute $\sup_{z\in\mathbb{R}}|g′(z)|$. It seemed to be alluded to with $f$. Or if you want to be clever and reuse f, then you can set $$g(x):=\delta f(\lambda x)=\frac{1}{3+x^4}$$ where you find that $\lambda =3^{-4}$ and $\delta=\frac{20}{3}$ so that $$\sup_{z\in\mathbb{R}}|g′(z)|=\delta\lambda \sup{z\in\mathbb{R}}|f′(z)|≤20(3^{-5})\frac{1}{5}<1$$ $\endgroup$ – Dayton Aug 21 at 21:49

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