Let $(f_n)$ sequence of differentiable functions on $[0,1]$ converging pointwise to $0$. Suppose
$|f'_n(x)| \leqslant 2015 + \cos(x)$$\forall x \in [0,1]$ and $\forall n$. Show that $(f_n)$ converge uniformly to $0$.
''Proof''
As supposed, $|f'_n(x)| \leqslant 2015 + \cos(x)$. We can rewrite it:
$|f'_n(x)| \leqslant 2015 + \cos(x) \leqslant 2016$ as $\cos(x)\leqslant1$
Since $(f_n)$ is a sequence of differentiable functions and the derivative is bounded, $(f_n)$ is Lipschitz. Hence, $\forall x,y \in [0,1]$:
$|f_n(x)-f_n(y)| \leqslant 2016|x-y|=\epsilon/3$
Moreover, as $(f_n)$ is Lipschitz, it implies that $(f_n)$ is uniformly continuous on $[0,1]$: $\forall \epsilon>0$$\exists \delta>0$ such that $\forall x,y \in [0,1]:$
$|x-y|<\delta \Rightarrow |f_n(x)-f_n(y)| \leqslant \epsilon/3$
And we want to show that $\forall \epsilon >0 \exists$$n_0$ such that $\forall n\geqslant n_0 \forall x \in [0,1]$:
$|f_n(x)|<\epsilon$.
Then i don't really see how to apply the pointwise convergence(i.e create kind a subdivision on $[0,1]$ or?) and conclude the proof using triangular inequality.
P.S Im stydying Analysis I right now so if you could explain with much details as possible it would be kind from your part. Thanks in advance ;)