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Let's consider the sequence of functions $f_n : [0,2] \to \mathbb{R} $ defined by $f_n(x)= (1+x^n)^{1/n} $ . I proved that this sequence converges pointwise to the function: $ f(x) = 1$ if $0\le x \le 1$ and $ f(x)=x$ otherwise. My problem is how can I prove that this sequence converges uniformly. (maybe I can prove that $f_n$ is a cauchy sequence but it look complicated) Please help me )=

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what did you try? did you try proving it straight out of the definition? those functions on [0,2] are bounded, try to use that. Also, did you prove the series is monotone decreasing for a given X? Try to prove the general case (uniform convergence occurs in all such cases of monotone decreasing sequences of functions) –  Benjamin Gruenbaum Oct 31 '12 at 7:12

1 Answer 1

  • If $0\leqslant x\leqslant1$, then $1\leqslant f_n(x)\leqslant2^{1/n}\leqslant1+1/n$.

  • If $x\geqslant1$, then $x\lt f(x)=x\cdot(1+x^{-n})^{1/n}\leqslant x\cdot(1+x^{-n}/n)\leqslant x+1/n$ since $n\geqslant1$.

This proves that $f(x)\leqslant f_n\leqslant f(x)+1/n$, hence $\|f_n-f\|_\infty\leqslant1/n\to0$. (The exact value being $\|f_n-f\|_\infty=2^{1/n}-1\sim\log2/n$, the order of this upper bound is the correct one.)

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I have a question! Why the second inequalities are useful? –  Daniel Oct 31 '12 at 10:17
    
@Daniel Without them nothing guarantees the convergence to $1$, respectively to $x$. –  Did Oct 31 '12 at 11:09
    
@Benjamin Gruenbaum, there is an easy way to prove that is monotone decreasing? –  Daniel Oct 31 '12 at 11:12
    
Thanks! :D It was useful –  Daniel Oct 31 '12 at 12:38
    
@Daniel yes, try using the binomial theorem and expanding the epxression. You can get a hint by first proving the easy case of (1+1/n)^n –  Benjamin Gruenbaum Oct 31 '12 at 16:43

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