Suppose $(f_n)$ is defined, continuous on $[a,b]$, and differentiable on this open interval. Then $c \in [a,b]$ and $(f_n(c))$ converge $(f'_n)$ and converge uniformly on $(a,b)$ respectively.

How can I prove that $(f_n)$ converges uniformly on $[a,b]$

I know by uniform cauchy criterion, $(f_n)$ converges uniformly on this set if we can find some $|f_n(x) - f_m(x)| \leq \epsilon$ for all $m, n \geq N$ and all $x \in a$

But we are only given that there is a single point $f_n(c)$ where it is uniformly convergent, nothing about the entire interval $[a,b]$

  • $\begingroup$ In fact even more is true: not only do the $(f_n)$ converge uniformly, the derivative of the uniform limit is the uniform limit of the $(f_n')$. $\endgroup$
    – Sarastro
    May 28 '15 at 19:27


$$f_n(x) - f_m(x) = [(f_n - f_m)(x) - (f_n - f_m)(c)]+(f_n - f_m)(c).$$

Use the MVT on the difference in brackets.

  • $\begingroup$ So we would get: $$ f'_n(c) - f'_m(c) = \frac{(f_n(x) - f_m(x)) - (f_n(c) - f_m(c)) + (f_n(c) - f_m(c))}{x-c}$$ Then, there is some $N \in \mathbb{N}$ such that $m,n \geq N$ $$|f'_n(c) - f'_m(c)| \leq \frac{\epsilon}{3}$$ Am I on the right track? $\endgroup$
    – elbarto
    May 28 '15 at 19:41
  • $\begingroup$ I'm not sure what you're doing. Don't we get $f_n(x) - f_m(x) = (f_n-f_m)'(d_x)\cdot (x-c) + (f_n-f_m)(c)?$ $\endgroup$
    – zhw.
    May 28 '15 at 20:56
  • $\begingroup$ zhigan, is that you? $\endgroup$
    – elbarto
    May 28 '15 at 22:25

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