# $(f_n)$ sequence of differentiable functions on $[0,1]$ and converge pointwise to $0$.

Let $$(f_n)$$ sequence of differentiable functions on $$[0,1]$$ converging pointwise to $$0$$. Suppose

$$|f'_n(x)| \leqslant 2015 + \cos(x)$$ $$\forall x \in [0,1]$$ and $$\forall n$$. Show that $$(f_n)$$ converge uniformly to $$0$$.

''Proof''

As supposed, $$|f'_n(x)| \leqslant 2015 + \cos(x)$$. We can rewrite it:

$$|f'_n(x)| \leqslant 2015 + \cos(x) \leqslant 2016$$ as $$\cos(x)\leqslant1$$

Since $$(f_n)$$ is a sequence of differentiable functions and the derivative is bounded, $$(f_n)$$ is Lipschitz. Hence, $$\forall x,y \in [0,1]$$:

$$|f_n(x)-f_n(y)| \leqslant 2016|x-y|=\epsilon/3$$

Moreover, as $$(f_n)$$ is Lipschitz, it implies that $$(f_n)$$ is uniformly continious on $$[0,1]$$: $$\forall \epsilon>0$$ $$\exists \delta>0$$ such that $$\forall x,y \in [0,1]:$$

$$|x-y|<\delta \Rightarrow |f_n(x)-f_n(y)| \leqslant \epsilon/3$$

And we want to show that $$\forall \epsilon >0 \exists$$ $$n_0$$ such that $$\forall n\geqslant n_0 \forall x \in [0,1]$$:

$$|f_n(x)|<\epsilon$$.

Then i don't really see how to apply the pointwise convergence(i.e create kind a subdivision on $$[0,1]$$ or?) and conclude the proof using triangular inequality.

P.S Im stydying Analysis I right now so if you could explain with much details as possible it would be kind from your part. Thanks in advance ;)

• How did the rewrite in the first line of your proof manage to slip in a "prime"? Jan 7, 2021 at 16:26
• @John Hughes fixed Jan 7, 2021 at 16:31
• The first appearance of $\epsilon$ in your proof does not make sense: What is meant by "for all $x,y \in [0,1]$ we have $2016|x-y| = \epsilon/3$"? The idea that should be pursued here is to chop $[0,1]$ into a grid of closely spaced points. Jan 7, 2021 at 16:52

Use @Michael's idea: first, to fix your observation about the uniform continuous of $$(f_n)$$ on $$[0,1]$$, you have to break your line

$$|f_n(x)-f_n(y)| \leqslant 2016|x-y|=\epsilon/3$$

into two separate statements:

1. $$(f_n)$$ is $$2016$$-Lipschitz.
2. $$\forall \epsilon > 0, \exists \delta > 0: \forall x,y \in [0,1], |x-y| < \delta \implies |f_n(x)-f_n(y)| < \epsilon / 3$$.

But #2 is still ambiguous since there's no quantifier and domain for the variable $$n$$ stated, and it's not clear if $$\delta$$ depends on $$n$$. To make things clear, you may fix $$\epsilon > 0$$ and define $$\delta = \epsilon / 6048$$, so that whenever $$|x-y| < \delta$$, #1 can be applied to get #2.

Then we proceed to the partition $$P = \{x_0 = 0,x_1 = \delta, x_2 = 2\delta, \dots, x_{M-1} = (M-1) \delta, x_M = 1\}$$, where $$M = \lceil 1/\delta \rceil$$ is the number of subintervals of $$P$$. Use pointwise convergence on each partition point $$x_i$$ to yield $$N_i \in \mathbb{N}$$ such that for all $$n \ge N_i$$, $$|f_n(x_i)| < \epsilon / 3$$.
Observation 2: for each partition point $$x_i$$, $$f_n(x_i)$$ is $$\epsilon$$-small when $$n$$ is sufficiently large.

Now go back to the paragraph about the uniform continuity of $$(f_n)$$. Apply this on a partition point $$x_i$$ and a (non-partition) point $$x \in [x_i,x_{i+1})$$, then $$|x_i - x| < \delta$$, so $$|f_n(x_i) - f_n(x)| < \epsilon / 3$$.
Observation 1: when $$x$$ is $$\delta$$-close to a partition point $$x_i$$, $$f_n(x_i)$$ approximates $$f_n(x)$$ $$\epsilon$$-closely, and the error $$\epsilon/3$$ is independent of $$n$$.

The question asks for the uniform convergence of $$(f_n)$$ on $$[0,1]$$, so we need $$N = \max\limits_{i \in \{0, 1, \dots, M\}} N_i$$, so that whenever $$n \ge N (\ge N_i)$$ and $$x \in [0,1]$$, $$|f_n(x)| \le |f_n(x) - f_n(x_i)| + |f_n(x_i)| \le \epsilon /3 + \epsilon + 3 < \epsilon.$$

To recap:

1. bounded derivatives gives subinterval length $$\delta$$ and $$(f_n)$$ Lipschitz.
2. use $$\delta$$ to build partition $$P = \{x_i\}_i$$.
3. apply pointwise convergence on partition points $$x_i$$. use $$N = \max_i N_i$$ to complete the logic.
4. use triangle inequality to conclude.

$$\require{AMScd}$$ $$\begin{CD} f_n(x_i) @<\delta \to 0< \text{uniform convergence} < f_n(x)\\ @V n \to \infty V \text{pointwise limit} V @.\\ 0 \end{CD}$$

• Thank you very much for clear and detailed explication! Jan 7, 2021 at 17:48