3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you mean "has a uniformly continuous derivative"? $\endgroup$ – Pedro Tamaroff Jun 18 '13 at 0:08
  • $\begingroup$ yes, sorry, my mistake $\endgroup$ – keri Jun 18 '13 at 0:16
0
$\begingroup$

What have you done so far?

Hint: for the pointwise convergence, write $$ f_n(x) = \frac{f(x+h_n)-f(x)}{h_n} $$ with $h_n=\frac{1}{n}\to 0^+$

$\endgroup$
  • $\begingroup$ Maybe it is better to write $h_n=1/n$. $\endgroup$ – Pedro Tamaroff Jun 18 '13 at 0:14
  • $\begingroup$ 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? $\endgroup$ – keri Jun 18 '13 at 0:29
  • 1
    $\begingroup$ 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 $g=f'$. $\endgroup$ – Clement C. Jun 18 '13 at 1:05
  • 1
    $\begingroup$ 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, $$ (currently writing) $\endgroup$ – Clement C. Jun 18 '13 at 1:49
  • 1
    $\begingroup$ In particular, $$ \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$$ $\endgroup$ – Clement C. Jun 18 '13 at 1:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.