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Show that there exists a unique continuous function $f : [0; 1] \rightarrow \mathbb{R}$ such that we have

$f(x) = \int^1_0\cos(x+f(y))dy$, for all $x \in [0, 1].$

Hint: use Banach fixed point theorem.

I defined a map $T:L[0,1] \rightarrow L[0, 1]$

$ Tf(x) = \dfrac{1}{3}\int^1_0\cos(x+f(y)dy$

and want to show that it is a contraction map, i.e. $d(Tf_1, Tf_2) < \alpha d(f_1, f_2), 0 < \alpha <1$, to apply Banach fixed point theorem, but do not know how to extract $f_1, f_2$ outside of the $cos(x + f(y))$.

Any hints or other help is appreciated.

Moisej Braver
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  • Use the mean-value theorem and the fact that the derivative of $\cos$ in absolute value is less or equal than 1. I would further suggest that you change the domain of $T$ and its codomain to $C([0,1],\mathbb{R})$. So that you can directly conclude that your fixed point is continuous. – Severin Schraven Mar 12 '17 at 18:00

1 Answers1

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Hint: Observe by the mean value theorem we have that \begin{align} |Tf(x)-Tg(x)| \leq& \frac{1}{3}\int ^1_0|\cos(x+f(z))-\cos(x+g(z))|\ dz\\ \leq&\ \frac{1}{3}\int^1_0|x+f(z)-(x+g(z))|\ dz = \frac{1}{3}\int^1_0|f(z)-g(z)|\ dz\\ \leq& \frac{1}{3}\sup_{z\in [0,1]}|f(z)-g(z)| \end{align} which means \begin{align} \|Tf-Tg\|_{\infty} \leq \frac{1}{3}\|f-g\|_\infty. \end{align}

I have assume $L[0, 1]$ means $L^\infty[0, 1]$.

Jacky Chong
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