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Let $f$ be a real valued, continuous and differentiable function on the real numbers with $f'$ uniformly continuous. Show that $g_n(x)$ with $$g_n(x)=nf(x+1/n)-nf(x)$$ is uniformly convergent to the derivative of $f$.

I cannot figure out how to solve this problem. If anyone could help me for this problem I would be very grateful.

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    Hint: $$g_n(x) = \frac{f(x+1/n)-f(x)}{1/n}$$ – 2017-01-04
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    "$g_n$ uniformly convergent to $f$" ?? Sure about that? – 2017-01-04
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    @Mark i cant understand please explain it. – 2017-01-04
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    If you write $g_n$ as I did, it's in the form of the [difference quotient](https://en.wikipedia.org/wiki/Difference_quotient), a common way to define the derivative. For this reason, it should be quite easy to show that $g_n\to f'$ (note that it's $f'$, not $f$) pointwise, so the only real work will to be to show that it does it uniformly. – 2017-01-04
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    @Mark i already show that $g_n(x)$ is pointwise convergent .but i cant get the inequality using derrivative of f is uniformly continuous. – 2017-01-04
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    @Did sorry for that, it is deerivative of f – 2017-01-04

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The key is to combine the uniform continuity of $f'$ with the mean-value theorem in order to deduce what is called "uniform differentiability." For any $h > 0$ we can use the MVT to find $0 < k < h$ such that $$ \frac{f(x+h) - f(x)}{h} - f'(x) = f'(x+k) - f'(x). $$

For $\epsilon >0$ we pick $\delta >0$ such that $z,y \in \mathbb{R}$ and $|z-y| < \delta$ implies that $|f'(z)-f'(y) | < \epsilon$. Then for $0 < h < \delta$ we use the above to estimate $$ \left\vert \frac{f(x+h) - f(x)}{h} - f'(x) \right\vert= |f'(x+k) - f'(x)| < \epsilon $$ since $0 < k < h < \delta$. Since $x$ is arbitrary we find that $$ 0 < h < \delta \Rightarrow \sup_{x } \left\vert \frac{f(x+h) - f(x)}{h} - f'(x) \right\vert < \epsilon. $$

This is actually a stronger result than what you need. I'll leave it to you to fit the two together.