2
$\begingroup$

Notation: $C([0,1],\mathbb{R})$ is the normed vector space of the continuous functions from $[0,1]$ to the real line, with norm $|f| = \sup \{|f(x)|, x \in [0,1]\}$

Show that there is a continuous funtcion $f:[0,1]\longrightarrow \mathbb{R}$ such that $\forall x \in [0,1]$ it satisfies:

$$f(x) = \frac{1}{3}\int_{0}^{x}t^2\sin(f(t))\,dt$$

Well, I am trying to solve this exercise. My idea is to show that the functional $J:C([0,1],\mathbb{R}) \rightarrow C([0,1],\mathbb{R})$ given by $J(f(x)) = \dfrac{1}{3}\int_{0}^{x}t^2\sin(f(t))\,dt$ has a fixed point. I am trying to show that this functional is a contraction and then conclude that there is a function $f$ such that $J(f) = f$, but I am struggling to find upper bounds for |J(f) - J(g)|. Any hints to find upper bounds for this functional?

$\endgroup$
5
  • 2
    $\begingroup$ Use that $|\sin(x)|\le|x|$. $\endgroup$ – Martín-Blas Pérez Pinilla Aug 22 '16 at 20:19
  • 1
    $\begingroup$ Martin's answer is the way to go . . . but if you explicitly want to find upper bounds for $|J(f) - J(g)|$, then $$ |J(f)- J(g)| = \left\vert \int_0^x t^2(\sin(f(t)) - \sin(g(t))) \; dt \right\vert \le \int_0^x t^2 | \sin f(t) - \sin g(t)| \; dt $$ and now apply Mean Value Theorem for $\sin f(t) - \sin g(t)$ $\endgroup$ – user288742 Aug 22 '16 at 20:24
  • 1
    $\begingroup$ The function $f(x)\equiv0$ is the only solution to this problem. $\endgroup$ – Omran Kouba Aug 22 '16 at 21:19
  • 1
    $\begingroup$ @user288742 Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – user99914 Aug 22 '16 at 23:29
  • 1
    $\begingroup$ Will do @ArcticChar $\endgroup$ – user288742 Aug 22 '16 at 23:34
4
$\begingroup$

Observe that $$|J(f(x))- J(g(x))| = \left\vert \int_0^x t^2(\sin(f(t)) - \sin(g(t))) \; dt \right\vert \le \int_0^x t^2 | \sin f(t) - \sin g(t)| \; dt $$ and by the Mean Value Theorem, $$\begin{split} & \sin f(t) - \sin g(t) = \cos(\xi_t) (f(t) - g(t)) \\ \implies &|\sin f(t) - \sin g(t)| \le |f(t) - g(t)| \le \|f-g\|_\infty \end{split}$$ so that $$|J(f(x))- J(g(x))| \le \|f - g\|_\infty \int_0^x t^2 dt \le \|f-g\|_\infty \int_0^1 t^2 \; dt = \frac{\|f-g\|_\infty}{3}. $$

Taking the supremum over all $x \in [0,1]$ on the LHS, we have $$ \|J f - Jg\|_\infty \le \frac{\|f-g\|_\infty}{3} $$ and by Contraction Mapping Theorem $J(f(x))$ has a unique fixed point.

$\endgroup$
1
  • 1
    $\begingroup$ It seems that $f(x)=2\arctan e^{x^3/9}$ : assume that $f'$ exists, then $\frac{f'(x)}{\sin f(x)} = \frac{x^2}{3}$, integrating both sides we get $f(x)=2\arctan e^{x^3/9}$. $\endgroup$ – Ignat Domanov Aug 25 '16 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.