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I saw 2 questions similar to mine, however I cannot seem to understand how to prove mine:

Uniformly bounded derivative implies uniform convergence

Uniform convergence, Bounded derivative.

The question: Let $f_n(x):[0,1] \to \mathbb R$ be a sequence of differentiable functions on $[0,1]$. Suppose:

  • $\lim_{n \to \infty} f_n(x)=f(x)$ exists $\forall x\in [a,b]$
  • The derivatives, $|f_n’|\le3$ $\forall n \in \mathbb N$.

Prove: $sup_x|f_n(x)-f(x)|\to 0$ as $n\to \infty$. Hint : use Heine–Borel theorem.

My thoughts for now:

  • prove that the sequence converge uniformly to $f(x)$ on a small open interval $(x-\delta,x+\delta)$ for all x
  • by using Heine-Borel theorem I can conclude that it happens on the whole $[0,1]$.

I don't understand how to start the first part.

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  • $\begingroup$ Isnt your question identical to the one in the linked posts? $\endgroup$ Commented Jun 26, 2021 at 18:03
  • $\begingroup$ It is very similar, I couldn't find a proof for the part I need. $\endgroup$ Commented Jun 26, 2021 at 18:11
  • $\begingroup$ It's identical, no? Can you point out, e.g., what is the difference between your question and those in the first link? $\endgroup$ Commented Jun 26, 2021 at 18:13
  • $\begingroup$ it is an identical question- the solution that i am looking for is different- the hint I have is Heine-Borel theorem. There is no recollection of that there $\endgroup$ Commented Jun 26, 2021 at 19:06

1 Answer 1

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Eventually I got some help and this is what I have:

Let $\epsilon>0$:

We know $[0,1]\subseteq\bigcup_{x\in[0,1]} B(x,\epsilon)$ is an open cover for the interval. Using Heine-Borel theorem, we know that $[0,1]$ is compact, and so, we have $I=\{1,...,k\}$ so $[0,1]\subseteq\bigcup_{i\in I} B(x_i,\epsilon)$ is a finite cover.

We have pointwise convergence, so for $i\in I \ \exists N_i\ \forall n,m\ge N_i\ : |f_n(x_i)-f_m(x_i)|<\epsilon$

We know, by using the mean value theorem that $\forall n\in \mathbb N\ \forall x<y\in [0,1] \ \exists c\in (x,y) : |\frac{f_n(y)-f_n(x)}{y-x}|=|f_n'(c)|\le 3$ $\Rightarrow$ $|f_n(y)-f_n(x)|\le 3\cdot |y-x|$

More specifically $\forall x\in B(x_i,\epsilon) : |f_n(x)-f_n(x_i)|\le 3\cdot |x-x_i|\le 3\cdot\epsilon$

Now, we choose $N= \max_{i\in I}\{N_i\}$. In conclusion:

$$\forall n,m \ge N \land \forall x\in [0,1] : |f_n(x)-f_m(x)|\le |f_n(x)−f_n(x_i)|+|f_n(x_i)−f_m(x_i)|+|f_m(x_i)−f_m(x)|<3\cdot \epsilon +\epsilon +3\cdot \epsilon =\epsilon '$$

We used Cauchy criteria for the sequence $\Rightarrow $ the sequence converges uniformly.

$\Downarrow $

$Sup_{x\in [0,1]}|f_n(x)-f(x)|\rightarrow 0 \ as\ n\rightarrow \infty $

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