Prove the existence and uniqueness of $\text{ } f(x) \in C^1([0,1])$ such that $\text{ }f: [0,1] \rightarrow [0,1]$ where \begin{equation} f(x) = 1 - (\int_{0}^{x} tf(t) dt)^2 \end{equation}
I was able to prove uniqueness using a standard argument of using continuity to show $f(x) - g(x) = 0$ on $x \in [0,1]$ is they satisfy the above integral equation. However, I'm stuck on figuring out how to show a solution exists.
My initial assumption was that one can show that a solution exists by using Banach fixed point theorem on $C^{1}[0,1]$, but I'm not sure how to bound the metric:
Let $T(f(x)) := 1 - (\int_{0}^{x} tf(t) dt)^2$, then \begin{equation} ||T(f(x)) - T(g(x))||_{\infty} = \sup_{x \in [0,1]} |(\int_{0}^{x} tg(t)) dt)^2 - (\int_{0}^{x} tf(t) dt )^2|\end{equation} But I'm not sure on how to bound the above expression to get a contraction mapping. An observation I made is the integral is an inner product $<t,f(t)>$, but when I wanted to use Cauchy Schwarz Inequality I realized I needed to use the triangle in the sup norm, from which I couldn't prove its a contraction mapping.
Any help on how to bound the sup norm or how else to proceed would be very helpful. Thank you!
A little algebra: \begin{align} \left(\int_0^x tgdt\right)^2-\left(\int_0^x tfdt\right)^2&=\left[\left(\int_0^x tgdt\right)-\left(\int_0^x tfdt\right)\right]\left[\left(\int_0^x tgdt\right)+\left(\int_0^x tfdt\right)\right]\\ &=\left(\int_0^x t(g-f)dt\right)\left(\int_0^x t(g+f)dt\right), \end{align} with $$ |\int_0^xt(f+g)dt|\le 2\int_0^xtdt\le 1, $$ and $$ |\int_0^xt(g-f)dt|\le \|g-f\|_\infty\int_0^xtdt\le\frac{1}{2}\|g-f\|_\infty. $$ This is enough for a contraction.