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Let $T: C([0,1]) \to C([0,1])$ be defined by $Tf(x) = x + \int_{0}^{x}tf(t)\,dt$. Prove that $T$ satisfies the hypotheses of the contraction mapping principle. Show that the fixed point is a solution to the differential equation $f'(x) = xf(x) + 1$.

How is my proof? I'm pretty iffy on it, though it seems right. I am specifically concerned about when I drop the sup term and then reintroduce it.

\begin{align*} d(Tf,Tg) &= \sup_{x\in [0,1]} \Biggl\lvert \biggl( x + \int_{0}^{x}tf(t)\,dt \biggr) - \biggl( x + \int_{0}^{x}tg(t)\,dt \biggr) \Biggr\rvert \\ &= \sup_{x\in [0,1]} \Biggl\lvert \int_{0}^{x} t\,\bigl(f(t)-g(t)\bigr)\,dt \Biggr\rvert \\ &\leq \sup_{x\in [0,1]} \int_{0}^{x} t\, \bigl\lvert f(t)-g(t) \bigr\rvert\,dt \\ &= \int_{0}^{1} t\, \bigl\lvert f(t)-g(t) \bigr\rvert\,dt \\ &\leq \int_{0}^{1} t\, \biggl(\, \sup_{t\in [0,1]} \lvert f(t)-g(t) \rvert \biggr) \, dt \\ &= \int_{0}^{1} t\, d(f,g) \,dt \\ &= d(f,g) \int_{0}^{1} t \,dt \\ &= \frac{1}{2} d(f,g) \end{align*} Now, suppose we have a fixed point $f$. We know that $$ T(f) = T\Bigl(\, \lim_{n \to \infty} T^n(f) \Bigr) = \lim_{n \to \infty} T^{n+1}(f) = f, $$ and so $$ f(x) = T(f(x)) = x + \int_{0}^{x}tf(t)\, dt \quad\implies\quad f'(x) = 1 + xf(x) $$ by the Fundamental Theorem of Calculus and the fact that $f$ is continuous.

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    $\begingroup$ @MartinR Fixed. $\endgroup$
    – beginner
    Dec 29, 2023 at 1:01

1 Answer 1

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All of the inequalities you have stated are correct.

One method to justify the inequality $$\sup_{x\in [0,1]} \int_{0}^{x}t|f(t)-g(t)|\,{\rm d}t = \int_{0}^{1}t|f(t)-g(t)|\,{\rm d}t$$ is to note that from the definition of the supremum, it is sufficient to show that $$\int_{0}^{x}t|f(t) - g(t)|\,{\rm d}t \leq \int_{0}^{1}t|f(t)-g(t)|\,{\rm d}t$$ for every $x\in [0,1]$.

Fix $x\in [0,1]$. Consider the function $\phi :[0,1] \to \mathbb{R}$ defined by $$\phi (t) := \begin{cases} t|f(t)-g(t)| & t\in [0,x], \\ 0 & t\in (x,1], \end{cases}$$ and the function $\psi : [0,1] \to \mathbb{R}$ defined by $$\psi (t) := t|f(t)-g(t)|.$$ Both $\phi$ and $\psi$ are integrable, either using the Riemann integral or the definition involving limits of step functions. As $t|f(t)-g(t)| \geq 0$ for every $t\in [0,1]$, it follows that $\phi (t) \leq \psi (t)$ for all $t\in [0,1]$. A direct consequence of this is $$\int_{0}^{1}\phi (t)\,{\rm d}t \leq \int_{0}^{1}\psi (t)\,{\rm d}t.$$ However, $$\int_{0}^{1}\phi (t)\,{\rm d}t = \int_{0}^{x}t|f(t)-g(t)|\,{\rm d}t$$ and $$\int_{0}^{1}\psi (t)\,{\rm d}t = \int_{0}^{1}t|f(t)-g(t)|\,{\rm d}t.$$ Therefore, $$\int_{0}^{x}t|f(t) - g(t)|\,{\rm d}t \leq \int_{0}^{1}t|f(t)-g(t)|\,{\rm d}t$$ as desired.

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  • $\begingroup$ Yes, the proof is correct. However, if you wanted to add some more details, I would suggest adding some explanation as to why that inequality holds, and a bit more explanation as to why $f(x) = x + \int_{0}^{x}tf(t)\,{\rm d}t$ implies $f'(x) = 1 + xf(x)$. $\endgroup$ Dec 30, 2023 at 1:12

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