# Proving existence and uniqueness of a fixed point for a function in a function space with a specific integral operator

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 $$[(f^*)'(x) = \frac{1}{2} \cos(f^*(x)) \quad \text{for all } x \in [0, 1].]$$

$$\mathbf{myidea(a)}$$

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 $$(b)$$

• b) is just FTC. Jun 28, 2023 at 6:17
• 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. Jun 28, 2023 at 6:25