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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.

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

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  • $\begingroup$ Ah great. I knew it was easy but couldn't see it. I appreciate the help. $\endgroup$ Commented Aug 2, 2023 at 19:31
  • $\begingroup$ Yes you were very close to the right answer. $\endgroup$
    – Kroki
    Commented Aug 2, 2023 at 19:32

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