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1. The problem statement, all variables and given/known data [math]\mbox{Check whether } \sum_{n=0}^\infty \frac {1}{e^{|x-n|}} \mbox{ is uniform convergent where its normaly convergent}[/math] 3. The attempt at a solution [math]\mbox{I choose } \epsilon = 1/2[/math] [math]a_n=\frac {1}{e^{|x-n|}}\ ,\ b_n= \frac {1}{n^2}[/math] [math]\lim_{n\rightarrow\infty} \frac {a_n}{b_n}=0\ \Rightarrow\ \sum_{n=0}^\infty a_n \mbox{ is convergent for all x }\in\ R. [/math] [math]f_k(x)=\sum_{n=0}^{k} a_n\ ,\ f(x)=\lim_{k\rightarrow\infty} \sum_{n=0}^{k} a_n[/math] [math] x_n=n\ \Rightarrow\ \sup_{x_n \in R}|f_k(x_n)-f(x_n)|\geq \sum_{n=k+1}^\infty \frac {1} {e^{|n-n|}}=1>\epsilon[/math] [math]\Rightarrow \mbox { is not uniformly convergent } \in R[/math] What do you think? [i left out technical details to present my idea more clearly.]

[math]\mbox{f(x) and f'(x) are continuous for all x in R. (*1)} [/math] [math]\mbox {Let a,b in R\ so\ that\ without\ the\ loss\ of\ generality } a<b.[/math] [math]\mbox {Let x in [a,b]. (*2)} [/math] [math]\lim_{n\rightarrow \infty} f_n(x)=\lim_{n\rightarrow \infty} n[f(x+\frac {1} {n})-f(x)]=[/math][math][t_n=1/n]=\lim_{t\rightarrow0} \frac {f(x+t)-f(x)} {t}=f'(x). \ (*3)[/math] [math]\mbox{(*1) and (*2) and (*3) } \Rightarrow\ g(x)=|f_n(x)-f(x)| \mbox{ is continuous for all x in R.\ (*4)}[/math] [math]\mbox{(*2) and (*4)} \Rightarrow\ \mbox {Weierstrass Theorem } \Rightarrow\[/math][math] \exists\ \max_{[a,b]} g(x)=g© \mbox { c in [a,b].\ (*5)} [/math] [math]\mbox {(*5) and (*1) } \Rightarrow\ \forall\ x\in[a,b]\ \Rightarrow\ \sup_{[a,b]}|f_n(x)-f(x)|=|\frac {f(c+t)-f©} {t} -f'©|\rightarrow0[/math] [math]\Rightarrow \mbox { Basic lemma for uniform convergence }[/math][math] \Rightarrow\ f_n(x) \mbox{ is uniform convergent in [a,b], where a,b are arbitrary.} [/math] QED [Edit] "Basic lemma for uniform convergence" is sounds little funny but I'm sure you'll understand what I meant. [Edit] I'm afraid of having the wrong intuition about uniform convergence, I hope you will be be able to say if I understand the idea from my proof. [Edit] If only I could prove g(x) is monotonic [such thing is possible?] then my proof would be a lot shorter, using Dini Theorem.

1. The problem statement, all variables and given/known data Suppose: [math]f(x)\ and\ f'(x)\ are\ continuous\ for\ all\ x \in R [/math] [math]For\ all\ x \in R\ and\ for\ all\ n \in N[/math] [math]f_n(x)=n[f(x+\frac{1}{n})-f(x)][/math] [math]Prove\ that\ when\ a,b\ are\ arbitrary,\ f_n(x)\ is\ uniform\ convergent\ in\ [a,b][/math] 3. The attempt at a solution [math]\lim_{n\rightarrow \infty} f_n(x)=\lim_{n\rightarrow \infty} n[f(x+\frac{1}{n})-f(x)] = \lim_{t\rightarrow0} \frac {f(x+t)-f(x)}{t}=f'(x)[/math] [math]\max_{[a,b]}|n[ f(x+\frac{1}{n}) -f(x)]-f'(x)|=|n[ f(x_0+\frac{1}{n}) -f(x_0)]-f'(x_0)|=[/math][math]|\frac {f(x_0+t)-f(x_0)}{t}-f'(x_0)|\rightarrow0[/math] I fear that I miss something terribly important. *I left out all the little technical details to make things shorter.