Exactly one solution to integral equations

Question

Show $\exists$ exactly one solution $U\in C([-1,1])$ to the intergral equation

$U(x)=x\int_{0}^{x}t^{2}cos(U(t)) dt $ for $x \in [-1,1]$

Attempt

I think I can use the contraction mapping theorem for this to prove there exists a unique fixed point and hence a unique soloution if I define $AU(x)=x\int_{0}^{x}t^{2}cos(U(t)) dt =U(x)$ so I want to show $\|{AU-AY}\|\leq k \|U-Y\|$ for $0<k\leq1$.

I get $\sup_{x\in[-1,1]}\bigg|x\int_{0}^{x}t^{2}(cos(U(t))-cos(Y(t)))\bigg|$

$\leq\int_{0}^{1}t^{2}(cos(U(t))-cos(Y(t)))dt$

I'm not sure how to get rid of the $t^{2}$ to get the results I want

Answer 1

I think that the norm $|| \cdot ||$ on $C([-1,1])$ is the $ \max - $ norm.

For $a,b \in \mathbb R$ we have $|\cos a- \cos b| \le |a-b|.$ This gives

$\int_{0}^{1}t^{2}|cos(U(t))-cos(Y(t))|dt \le \int_{0}^{1}t^{2}|U(t)-Y(t)|dt$

$ \le \int_{0}^{1}t^{2}||U-Y||dt= ||U-Y||\int_{0}^{1}t^{2} dt= \frac{1}{3}||U-Y||.$

Answer 2

$$ \Vert Tu - Tv \Vert_{\infty} =\sup_{x \in [-1,1]} \vert Tu(x)-Tv(x) \vert \le \int_0^1 t^2 \vert \cos u(t)- \cos v(t)\vert \, dt $$ $$ \le \Vert u - v \Vert_{\infty} \int_0^1 t^2 \, dt = \frac{1}{3} \Vert u-v \Vert_{\infty} $$ being the cosine function $1$-Lipschitz. Then you conclude by Contraction fixed point Theorem.