# Banach fixed point theorem to solve integral equation

Let $$\alpha>0$$. I need to show there is a unique real-valued continuous function $$f$$ on [0,1] such that $$f(x) = \int_{0}^xt^{\alpha}f(t)dt$$ for all $$x \in [0,1]$$.

I am pretty confident using Banach's fixed point theorem is the way to go here. I defined $$T:C([0,1]) \to C([0,1])$$ as $$T(f)(x) = \int_{0}^xt^{\alpha}f(t)dt.$$ Since $$C([0,1])$$ with the supremum norm ($$\lVert f\rVert_{\infty} = \sup_{x \in [0,1]}|f(x)|$$) is complete, I only have to show $$T$$ is a contraction mapping.

To do this, let $$f,g \in C([0,1])$$. Then \begin{align} \lVert T(f)(x) - T(g)(x)\rVert_{\infty} &= \left\lVert \int_{0}^xt^{\alpha}(f(t)-g(t))dt\right\lVert_{\infty}, \end{align} but from here I struggle to find the correct bounds to show $$T$$ is a contraction. I know that \begin{align} \int_{0}^xt^{\alpha}(f(t)-g(t))dt &\leq x \sup_{t \in [0,x]}|t^{\alpha}(f(t)-g(t))| \\ &\leq x^{\alpha+1}\sup_{t \in [0,x]}|f(t)-g(t)| \\ &\leq x^{\alpha+1}\sup_{t\in [0,1]}|f(t) - g(t)| \end{align} but since $$x \in [0,1]$$, $$x^{\alpha+1}$$ is maximized at $$x=1$$, which doesn't work as a contraction constant.

I'm thinking there is a clever way to get an upper bound that is strictly less than 1, but I can't find it. I'd appreciate a hint instead of a direct answer if that is possible, but if I am completely on the wrong path for the problem, that would be good to know too.

$$\left\|T(f) - T(g)\right\|_{\infty} \le \left\|f - g\right\|_{\infty}\sup_{x\in [0,1]}\int_0^x t^\alpha \mathrm d t = \frac1{\alpha + 1}\left\|f - g\right\|_{\infty}$$