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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.

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

1 Answer 1

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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 $h_n=\frac{1}{n}\to 0^+$

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  • $\begingroup$ Maybe it is better to write $h_n=1/n$. $\endgroup$
    – Pedro ♦
    Jun 18, 2013 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, 2013 at 0:29
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    $\begingroup$ 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, 2013 at 1:05
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    $\begingroup$ 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, $$ (currently writing) $\endgroup$
    – Clement C.
    Jun 18, 2013 at 1:49
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    $\begingroup$ In particular, $$ \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, 2013 at 1:57

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