Hi I was having trouble with the following question:
A user called Fischer has said that "The convergence is uniform on every compact subset of $\mathbb{R}$, however", without providing a proof. (Uniform convergence to the derivative)
Help will be appreciated?
Note that by the mean value theorem,
$$ \sup_{x \in [\alpha, \beta]} |g_n(x) - f'(x)| = \sup_{x \in [\alpha, \beta]} \left| n \left( f \left(x + \frac{1}{n} \right) - f(x) \right) - f'(x) \right| = \sup_{x \in [\alpha, \beta]} |f'(c(x,n)) - f'(x)|$$
where $c(x,n)$ is some point between $x$ and $x + \frac{1}{n}$. Choose $N$ such that $\left[ \alpha - \frac{1}{N}, \beta + \frac{1}{N} \right] \subset (a,b)$ and let $\varepsilon > 0$. Since $f'$ is continuous on the closed interval $\left[ \alpha - \frac{1}{N}, \beta + \frac{1}{N} \right]$, it is uniformly continuous there and so we can find $\delta > 0$ such that if $x,y \in \left[ \alpha - \frac{1}{N}, \beta + \frac{1}{N} \right]$ and $|x - y| < \delta$ then $|f'(x) - f'(y)| < \varepsilon$. Hence, if $n > \max \{ N, \frac{1}{\delta} \}$ and $x \in [\alpha, \beta]$, we have $c(x,n) \in \left[ \alpha - \frac{1}{N}, \beta + \frac{1}{N} \right]$ and $|c(x,n) - x| < \frac{1}{n} < \delta$ and thus
$$ \sup_{x \in [\alpha, \beta]} |f'(c(x,n)) - f'(x)| < \varepsilon. $$