# Suppose $f_n : [0,1]\rightarrow\mathbb{R}$ is a sequence of $C^1$ functions that converges pointwise to $f$.

I came across this problem in Davidson's real analysis text and wanted some input on whether my thinking is valid. This is section 8.1, problem H. Here is the full statement of the problem.

Suppose that $f_n : [0,1]\rightarrow \mathbb{R}$ is a sequence of $C^1$ functions that converges pointwise to a function $f$. If there is a constant $M$ so that $||(f_n)^{\prime}||_\infty\leq M$ for all $n$, prove that $(f_n)$ converges uniformly.

Here is my attempt at a solution:

Take $\epsilon > 0$ and let $x_0\in(0,1)$. Since $f_n$ converges to $f$ pointwise, there exists an $N\in\mathbb{N}$ such that for all $n\ge N$, $|f_n(x_0) - f(x_0)| < \frac{\epsilon}{2}$. Now, since $|f^{\prime}_n|\leq M$, if $\delta > 0$, we can write $|f_n(x_0 + \delta)|\leq|f_n(x_0)| + M\delta$.

Here is where I think I'm taking an unwarranted leap. I want to say that the condition on $f^{\prime}_n$ also implies that $|f(x+\delta) - f(x)|\leq M\delta$, but I'm not sure that I can justify this.

However, if this is true, we can continue and say that if $x\in (x_0-\delta,x_0+\delta)$, then $|f_n(x)-f(x)|\leq |f_n(x_0)-f(x_0)| + 2M\delta$ and we can take $\delta < \frac{\epsilon}{4M}$ such that $|f_n(x) - f(x)| < \epsilon$. This implies we can cover the interval [0,1] with $m$ open balls of radius $\frac{\epsilon}{4M}$ and in each case get some $N_k$ such that for all $n\geq N_k$, $|f_n(x) - f(x)| < \epsilon$ for $x$ in that respective ball. Thus, we can take $N = \text{max}\{N_1,N_2,\dots,N_m\}$ so that for all $x\in[0,1],\ |f_n(x) - f(x)| < \epsilon$ for all $n\geq N$.

Does this work? is there a simpler way to look at this problem?

And tend $n$ to infinity. To do it more properly \begin{align*} |f_n(x) - f(x)| &\le |f_n(x) - f_n(k\delta)| + |f_n(k\delta) - f(k\delta)| + |f(k\delta) - f(x)| \\& \le 2M|x-k\delta| + |f_n(k\delta) - f(k\delta)| \end{align*}
For every $k \in \mathbb{N}$. By taking the minimum over all $k$ such that $k \le \frac{1}{\delta} + 1$. We have :