Prove that a sequence of recursive functions $\,f_n(x)$ cannot converge pointwise to $\,f(x)$ on $[0,1]$

Question

Given a recursive sequence $\,f_n(x) :[0,1] \to \mathbb R$, $x \in [0,1]$, where $$\begin{align*} f_1(x) &= x, \\[6pt] f_n(x) &= \frac{2x\,f_{n-1}(x)}{n!} \end{align*}$$ I have proven that the sequence is non-negative and non-increasing, but cannot prove that its pointwise convergent. I'm not even sure if it is pointwise convergent! Im having trouble defining an $N(x,\epsilon)$ to use for the Cauchy definition for a convergent sequence. The recursion is really throwing me off.

Answer 1

Hint: for any $x\in[0,1]$,

Using a recursion $$ \begin {cases} f_1(x) = 1 =: u_1 \\ f_{n+1}(x) = \frac{2}{n!}f_n(x) \le \frac{2}{n!}u_n =: u_{n+1} \end {cases} $$

so it is enough to prove that $u_n \to 0$.

Answer 2

It is given that $f_1(x)=x$ and $f_n(x)=\frac{1}{n!}2x f_{n-1}(x)$. In particular, for every $x\in[0,1]$ and $n\geq3$: $$0\leq f_n(x)=\frac{1}{n!}2x f_{n-1}(x)\leq\frac{1}{n!}2 f_{n-1}(x)\leq\frac{1}{2}f_{n-1}(x)$$

Suppose that $f_3(x)\leq M$ for every $x\in[0,1]$. Then $f_{3+k}(x)\leq \frac{M}{2^k}$. Do you understand that this implies uniform convergence?