Let $f_{n}(x)$ $=$ $n^{2}x\ (1-x)^{n}$ with $0\le x\le 1$, where $f_{n}$ converges pointwise to the zero function $f$.
How do I check for uniform convergence?
Can someone provide me with some hints?
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1Say something about what you have tried. – eccstartup Jul 26 '13 at 01:42
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Try this technique. – Mhenni Benghorbal Jul 26 '13 at 01:55
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1People tend to downvote if you don't include some of your attempted work; I recommend you to put some of your work down, that way people would be more likely to help ... – Jul 26 '13 at 01:57
3 Answers
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Convergence is not uniform. $$f_n(x)=n^2x(1-x)^n$$
Let $x=n^{-1}$. Then $$f_n(n^{-1})=n\left(1-\frac 1 n\right)^n$$ is unbounded as $n\to\infty$.
Thus convergence cannot be uniform.
Pedro
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Use the theorem for uniform convergence and Riemann Integrability.
Integrate each of your functions of $\{f_n\}$ in $[0,1]$ and take limit $n \rightarrow \infty$. You shall get 1. Whereas you shall get 0 after integrating the limit function.
Supriyo
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Hint: Calculate $\sup\limits_{0\leq x\leq 1}\left|f_n(x)-f(x)\right|$
(edit: that's literally common sense, so adding one more line)
$$\sup_{0\leq x\leq 1}|f_n(x)-f(x) |=\left(1-\frac1{n+1}\right)^n\times\frac{n^2}{n+1}$$