Question: Let $\mathcal{X} = C([0, 1], \mathbb{R})$ (the space of real-valued continuous functions on the closed interval $[0, 1]$) with the uniform metric $d(f, g) = \max_{x \in [0, 1]} |f(x) - g(x)|$. Define $T: \mathcal{X} \to \mathcal{X}$ by
$$T(f(x)) = 1 + \int_0^x tf(t) dt, 0 \leq x \leq 1.$$
- Prove that $T$ is a contraction map.
- Let $f_0(x) = 1$ for all $x \in [0, 1]$. Set $f_{k + 1}(x) = T(f_k(x))$, for all $k \geq 0$. Find explicitly $f_n(x)$ for $n = 1, 2, 3$. Then prove that the sequence $\{f_n(x)\}$ converges in $(\mathcal{X}, d)$ to $f(x) = e^{x^2/2}$, for all $x \in [0, 1]$.
Answer:
I was able to prove that the mapping $T$ is a contraction map with Lipschitz constant 1/2.
This is where I am confused. Let me present my work first.
It follows that
$$f_1(x) = T(1) = 1 + \int_0^x t \cdot 1 dt = 1 + \frac{x^2}{2},$$
$$f_2(x) = T(1 + x^2/2) = 1 + \int_0^x t\cdot\bigg(1 + \frac{x^2}{2}\bigg)dt = 1 + \frac{x^2}{2} + \frac{x^4}{4},$$
and
$$f_3(x) = T(1 + x^2/2 + x^4/4) = 1 + \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{8}.$$
In general, we then have that
$$f_n(x) = \sum_{k = 0}^{n} \bigg(\frac{x^2}{2}\bigg)^k.$$
Thus,
$$\lim_{n \to \infty}f_n(x) = \frac{1}{1 - \big(\frac{x^2}{2}\big)} = f(x).$$
So, did I mess up or is this problem asking the wrong question?