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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 ;)

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  • $\begingroup$ How did the rewrite in the first line of your proof manage to slip in a "prime"? $\endgroup$ Jan 7, 2021 at 16:26
  • $\begingroup$ @John Hughes fixed $\endgroup$
    – Daniil
    Jan 7, 2021 at 16:31
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    $\begingroup$ 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. $\endgroup$
    – Michael
    Jan 7, 2021 at 16:52

1 Answer 1

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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}

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  • $\begingroup$ Thank you very much for clear and detailed explication! $\endgroup$
    – Daniil
    Jan 7, 2021 at 17:48

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