I am trying to solve this problem:
For $x\in[-1,1],$ consider the equation $$[1+\epsilon\sin(f(x))]\cdot f(x)=x^2$$for the unknown function $f.$ Use the contraction mapping theorem to show that there exists an $\epsilon_0>0$ such that, for each $0\leq\epsilon<\epsilon_0,$ the equation above has a unique solution $f\in C([-1,1]).$
In my attempt to show this, I first expanded the equation above to read $$f(x)+\epsilon f(x)\sin(f(x))=x^2$$ $$\implies f(x)=x^2-\epsilon f(x)\sin(f(x)),$$so I set $\Phi:C([-1,1])\rightarrow C([-1,1])$ as $$\Phi[f](x)=x^2-\epsilon f(x)\sin(f(x)).$$I am also equipping $C([-1,1])$ with the metric $d_\infty$ where $||f||_\infty=\sup_{x\in[-1,1]}|f(x)|.$
I first looked at $|\Phi[f]-\Phi[g]|,$ and cancelled the $x^2$ terms in each expression to get $$=\epsilon|g(x)\cdot\sin(g(x))-f(x)\cdot\sin(f(x))|.$$However, I'm getting a bit stuck here on how to proceed and eventually get $||\Phi[f]-\Phi[g]||_\infty\leq C||f-g||_\infty.$ I figure that I'll need to eventually use mean value theorem on $\sin(\hspace{0.1cm}\cdot\hspace{0.1cm}),$ but I'm not sure how to get there with the $g(x)$ and $f(x)$ in the way in that expression, and also how $\epsilon_0$ will be introduced.
Any guidance is appreciated.
Thank you.
I am not sure if your approach works. The problem is that the function $h(w) = w \sin(w)$ is not Lipschitz continuous on $\Bbb R$, which makes it difficult to estimate $$ |g(x)\cdot\sin(g(x))-f(x)\cdot\sin(f(x))| = |h(g(x)) - h(f(x))| $$ in terms of $|f(x) - g(x)|$.
But you can define the contraction mapping as $$ \Phi[f](x) = \frac{x^2}{1+\epsilon\sin(f(x))} \, , $$ this is defined if $0 \le \epsilon < 1$. Then $$ |\Phi[f](x) - \Phi[g](x)| = x^2 \left| \frac{1}{1+\epsilon\sin(f(x))} - \frac{1}{1+\epsilon\sin(g(x))}\right|\\ \le \left| \frac{1}{1+\epsilon\sin(f(x))} - \frac{1}{1+\epsilon\sin(g(x))}\right| \\ = \frac{\epsilon |\sin(f(x) - \sin(g(x)|}{(1+\epsilon\sin(f(x)))(1+\epsilon\sin(g(x)))} \, . $$ The sine function is Lipschitz continuous with Lipschitz constant equal to one, therefore $$ |\Phi[f](x) - \Phi[g](x)| \le \frac{\epsilon}{(1-\epsilon)^2} |f(x) - g(x)| \, . $$ It only remains to show that the factor $\frac{\epsilon}{(1-\epsilon)^2}$ is strictly less than one for sufficiently small $\epsilon$.