# Pointwise convergence and uniform convergence

Suppose that $f$ has a uniformly continuous derivative. We define $\ f_n: \Bbb R\to\Bbb R$ by

$$\ f_n(x) = n \left( f \left(x + \frac{1}{n}\right) - f(x)\right)$$

Find a pointwise convergence $\ f_n$. Prove that the sequence $\ f_n$ converges uniformly to its limit.

• Do you mean "has a uniformly continuous derivative"?
– Pedro
Jun 18, 2013 at 0:08
• yes, sorry, my mistake
– keri
Jun 18, 2013 at 0:16

Hint: for the pointwise convergence, write $$f_n(x) = \frac{f(x+h_n)-f(x)}{h_n}$$ with $h_n=\frac{1}{n}\to 0^+$
• Maybe it is better to write $h_n=1/n$.
• so pointwise convergence: ∀x∈D:∀ϵ∈R>0:∃N∈R:∀n>N:|fn(x)−f(x)|<ϵ $f_n(x) = \frac{f(x+\frac{1}{n})- \frac{n+1}{n} f(x)}{\frac{1}{n}}$ What should I do next?
• Convergence pointwise of $(f_n)$ means that there exists a function $g$ such that, for every fixed $x_0$, $$f_n(x_0)\xrightarrow[n\to\infty]{} g(x_0)$$ Here, for any $x_0\in\mathbb{R}$, you have $$f_n(x_0)=\frac{f(x_0+1/n)-f(x_0)}{1/n}\xrightarrow[n\to\infty]{} f'(x_0)$$ (by definition of the derivative of $f$) proving that $(f_n)$ converges pointwise to $g=f'$. Jun 18, 2013 at 1:05
• For the second part, use the mean value theorem. More previsely, fix any $\epsilon > 0$; and let $\eta$ be the corresponding value wrt the uniform convergence of $f'$. Further, let $N$ be the first integer such that $\frac{1}{N} \leq \eta$. We hereafter consider only integers $n\geq N$. For any $x$ in $\mathbb R$, there exists $c_n=c_n(x)$ such that $\frac{f(x+1/n)-f(x)}{1/n}=f'(c_n)$ (mean value theorem). Thus, $f_n(x) = f'(c_n)$. In particular, $$(currently writing) Jun 18, 2013 at 1:49 • In particular,$$ \forall n\geq N,\forall x,\ |f_n(x)-f'(x)|=|f'(c_n(x))-f'(x)| \leq \epsilon $$because |c_n(x)-x|\leq 1/n\leq 1/N \leq \eta. This shows that \forall \epsilon > 0, \exists N_\epsilon, \forall n\geq N_\epsilon \sup_x |f_n(x)-f'(x)| \leq \epsilon; in other terms,$$\sup_{x\in\mathbb{R}} |f_n(x)-f'(x)| \xrightarrow[n\to\infty]{} 0 Jun 18, 2013 at 1:57