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 August 3rd, 2012, 07:46 AM #1 Newbie   Joined: Jul 2012 Posts: 2 Thanks: 0 Show that a continuous derivative is a uniform limit Suppose $f$ is differentiable on $[a,b]$, where the derivative at an endpoint is the one-sided derivative, and suppose that the derivative is continuous. Show that the sequence of functions $g_{n}:[a,b]\rightarrow\mathbb{R}$$g_{n}(x)=n\left(f\left(x+\frac{1}{n}\right)-f(x)\right),\, n=1,\,2,\,\ldots$ converges uniformly to $f'(x)$ on any compact subinterval of $[a,b)$.
 August 3rd, 2012, 05:27 PM #2 Senior Member   Joined: Jul 2011 Posts: 118 Thanks: 0 Re: Show that a continuous derivative is a uniform limit $\lim_{n\to\infty}g_n(x)=\lim_{n\to\infty}\frac{f$$x+\frac{1}{n}$$-f(x)}{\frac{1}{n}}=\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=f#39;_+(x)$ so the limit exists in $[a,b)$ Using mean value theorem: $\|f'(x)-g_n(x)\|=\|f'(x)-n$$f\(x+\frac{1}{n}$$-f(x)\)\|=\|f'(x)-nf'$$x+\theta(x,n)\cdot\frac{1}{n}$$\cdot\frac {1}{n}\|=\|f'(x)-f#39;$$x+\theta(x,n)\cdot\frac{1}{n}$$\|,\ 1>\theta(x,n)>0$ This is valid only for any interval $[a,c]\subset[a,b)$ as $x+\frac{\theta(x,n)}{n}$ must lie in $[a,b]$ so that derivative exists As $f'(x)$ is continuous in closed interval $[a,b]$ it is uniformly continuous there. So for any $\delta>0$ we can find $\epsilon>0$ such that if $\frac{\theta(x,n)}{n}<\frac{1}{n}<\epsilon$ then $\|f'(x)-f'$$x+\frac{\theta(x,n)}{n}$$\|<\delta$. Thus for $n>\frac{1}{\epsilon}$ we get $|f'(x)-g_n(x)|<\delta$ for all $x\in[a,c]$

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