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It's right this proof?

Let $(f_n)_n$ a sequence of differentiable functions on $[a,b]$, such that $f_n$ converges pointwise to f and $\forall x \in [a,b] \exists M>0 : |f_n ' (x)| \le M \ \forall n \in \mathbb{N}$.

Then $f_n$ is uniformly convergent to $f$.

I estimate $|f(x)-f_n(x)|$

Let $y \in [a,b]; f_n(x)=f_n (y) + f_n ' (\xi_n) (x-y)$

$f(x)=f(y)+f'(\xi) (x-y)$

Then: $|f(x)-f_n(x)|=|f(y)+f'(\xi) (x-y)-(f_n (y) + f_n ' (\xi_n) (x-y)|$

$\le|f(y)-f_n(y)|+|f ' (\xi)-f_n ' (\xi_n)||x-y|\le|f(y)-f_n(y)|+2M|x-y|$

Let $\epsilon >0$ then $\exists \delta >0 : |x-x_0|<\delta \Rightarrow |f_n(x)-f(x_0)|<\epsilon$

I choose $y$ such that $|x-y|<min({\delta, \epsilon})$

Then $|f(x)-f_n (x)|\le(1+2 M) \ \epsilon$

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And what exactly is your question? This isn't a question. – martini Nov 29 '12 at 16:02
It's right?.... – Madara Nov 29 '12 at 16:03
You should write this explicitly above. And: Where do you get the differentiability of $f$ from? You used $f'$ in your argument ... Modify your argument with $f_m$ replacing $f$, show that $(f_n)$ is uniformly cauchy ... And: Your $y$ do not work for all $x$ .. but you want uniform convergence ... – martini Nov 29 '12 at 16:08

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