Let $g_n(x)=n[f(x+\frac{1}{n})-f(x)]$, where $f: R\to R$ is a continuous function. Which of the following are true?

a. If $f(x)=x^3$, then $g_n\to f'$ uniformly on $R$ as $n\to \infty$.

b. If $f(x)=x^2$, then $g_n\to f'$ uniformly on $R$ as $n\to \infty$.

c. If f is differentiable and if $f'$ is uniformly continuous on $R$, then $g_n\to f'$ uniformly on $R$ as $n \to \infty$.

Here For (a) , $g_n(x)=n[(x+\frac{1}{n})^3-x^3]=n[\frac{1}{n^3}+\frac{3x^2}{n}+\frac{3x}{n^2}]=\frac{1}{n^2}+3x^2+\frac{3x}{n}$. Now $g_n(x)\to f'$ for sure . For uniform convergence $sup|g_n(x)-f'(x)|=sup|\frac{1}{n^2}+\frac{3x}{n}|$. This doesn't converge as the max value is unbounded.

For (b) $g_n(x)=n[(x+\frac{1}{n})^2-x^2]=n[\frac{2x}{n}+\frac{1}{n^2}]=2x+\frac{1}{n}$. Now

Now $g_n(x)\to f'$ for sure. For uniform convergence part $|g_n(x)-f'(x)|=\frac{1}{n}\lt \epsilon$ . hence (b) is true.

Now for (c), since (a) is false, (c) is false too.

Am I right??

  • $\begingroup$ @jim How so??.. Will not that contradict (a)?? $\endgroup$ Jan 26 '14 at 5:59


where $x_n\in(x,x+\frac{1}{n})$

then use uniform continuity of $f^{\prime}$ to get the result.

  • $\begingroup$ How U could write that? I can not understand.. Can you please elaborate??@jim $\endgroup$
    – user450210
    Oct 3 '17 at 19:17
  • $\begingroup$ How could I tell $f'$ is uniform continuity? $\endgroup$
    – user464147
    Oct 12 '17 at 17:14

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