Consider $X = C^0([0, 1], \mathbb{R})$ equipped with the $\|\cdot\|_{\infty}$ norm, and let $T : X \rightarrow X$ be defined as $[T(f)(x) = \frac{1}{2} \int_{0}^{x} \cos(f(t)) \, dt.]$ We assume that $T$ is well-defined.

(a) Show that $T$ is a contraction and conclude that $T$ has a unique fixed point $f^* \in X$, i.e., $[f^*(x) = \frac{1}{2} \int_{0}^{x} \cos(f^*(t)) \, dt \quad \text{for all } x \in [0, 1].]$

(b)Conclude that $f^* \in C^1([0, 1], \mathbb{R})$ with $f^*(0) = 0$ and $[(f^*)'(x) = \frac{1}{2} \cos(f^*(x)) \quad \text{for all } x \in [0, 1].]$


To show that $T$ is a contraction, we need to prove the Lipschitz condition. That is, we need to find a constant $0 \leq k < 1$ such that for all $f, g \in X$, we have First, let's consider $|T(f) - T(g)\|_\infty$: $[|T(f) - T(g)\|_\infty = \sup_{x \in [0, 1]} |T(f)(x) - T(g)(x)|.]$ Now, let's substitute the definition of $T$: \begin{align*} \|T(f) - T(g)\|_\infty &= \sup_{x \in [0, 1]} \left|\frac{1}{2}\int_0^x \cos(f(t))\,dt - \frac{1}{2}\int_0^x \cos(g(t))\,dt\right|. \end{align*} Using the triangle inequality, we have: \begin{align*} \|T(f) - T(g)\|_\infty &\leq \frac{1}{2} \sup_{x \in [0, 1]}\left|\int_0^x \cos(f(t))\,dt - \int_0^x \cos(g(t))\,dt\right| \\ &= \frac{1}{2} \sup_{x \in [0, 1]}\left|\int_0^x \left(\cos(f(t)) - \cos(g(t))\right)\,dt\right|. \end{align*} Since the function $\cos(t)$ is continuous, we have for every $t$: [|\cos(f(t)) - \cos(g(t))| \leq |f(t) - g(t)|.] Hence, we can further estimate the above inequality as follows: \begin{align*} \|T(f) - T(g)\|_\infty &\leq \frac{1}{2} \sup_{x \in [0, 1]}\left|\int_0^x |f(t) - g(t)|\,dt\right| \\ &\leq \frac{1}{2} \sup_{x \in [0, 1]} \left|x\int_0^1 |f(t) - g(t)|\,dt\right| \\ &\leq \frac{1}{2} \sup_{x \in [0, 1]} \left|x \cdot \|f - g\|_\infty\right| \\ &= \frac{1}{2} \cdot x_0 \|f - g\|_\infty, \end{align*} where $x_0$ is a constant satisfying $0 \leq x_0 \leq 1$. To guarantee the Lipschitz condition, we need to find a constant $k$ such that $\frac{1}{2} \cdot x_0 < k < 1$. Since $0 \leq x_0 \leq 1$, it follows that $\frac{1}{2} \cdot x_0 < 1$. Therefore, $T$ is a contraction. By the Banach fixed-point theorem, we know that $T$ has a unique fixed point $f^* \in X$. This means there exists a function $f^*$ such that $[f^*(x) = \frac{1}{2}\int_0^x \cos(f^*(t))\,dt,]$ for all $x \in [0, 1]$. Thus, we have shown that $T$ has a unique fixed point $f^* \in X$ satisfying the given equation.

I'm not sure about my idea to the part $(a)$ and I'm still not having an appropiate idea to the part $(b)$

  • 1
    $\begingroup$ b) is just FTC. $\endgroup$ Jun 28, 2023 at 6:17
  • $\begingroup$ Your reasoning for $(a)$ seems correct, though I think you meant to say "since the function $\cos$ is $1$-Lipschitz" instead of just continuous, but other than that it's fine I believe. $\endgroup$
    – Bruno B
    Jun 28, 2023 at 6:25


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