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How to prove uniform convergence for sequences $f_n = x^n(1-x), f_n = \frac {n^3x} {1+n^4x^2}$ and $ f_n = \sqrt {n} xe^{-nx^2}$ on $[0,1]$

I've already shown that the following sequence of functions converge pointwise to $0$ on $[0,1]$:

$$f_n = \frac {2x} {1+nx}$$

$$f_n = x^n(1-x)$$

$$f_n = \frac {n^3x} {1+n^4x^2}$$

$$f_n = \sqrt {n} xe^{-nx^2}$$

For $f_n = \frac {2x} {1+nx}$ I have $\frac {2x} {1+nx} =\frac {2} {1/x+n} \le \frac 2 n \le \epsilon$, so I can choose $N$ large enough and the inequality will hold for all $x \in [0,1]$.

However, I don't see how to prove whether or not the other three functions converge uniformly. I'm not sure I'm supposed to use majorant series, since this exercise is a basic one.

Could someone point out how to prove uniform convergence for the three sequences in question ?

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1 Answer 1

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I will show the 1st function $f_n(x) = x^n\cdot (1 - x)$ to converge uniformly on $[0, 1]$. $f_n'(x) = nx^{n-1} - (n+1)x^n = 0 \iff x = 0, \dfrac{n}{n+1}$. So $\displaystyle \sup_{x \in [0, 1]} |f_n(x) - 0| = f_n\left(\frac{n}{n+1}\right) = \dfrac{1}{n+1}\cdot \dfrac{1}{(1 + \frac{1}{n})^n} \to 0$ as $n \to \infty$.

For the second function $f_n(x) = \dfrac{xn^3}{1 + n^4x^2}$, we have: $f_n'(x) = \dfrac{n^3 - x^2n^7}{(1 + n^4x^2)^2} = 0 \iff n^3 - x^2n^7 = 0 \iff x = \dfrac{1}{n^2}$. So $\displaystyle \sup_{x \in [0,1]} |f_n(x) - 0| = f_n\left(\frac{1}{n^2}\right) = \dfrac{n}{1 + n^2} \to 0$ as $n \to \infty$.

Surprisingly, the 3rd function does not converge uniformly on $[0, 1$. To see this, $f_n(x) = \sqrt{n}xe^{-nx^2}$ has $f_n'(x) = \sqrt{n}e^{-nx^2}\cdot \left(1 - 2nx^2\right) = 0 \iff x = \dfrac{1}{\sqrt{2n}}$. So $\displaystyle \sup_{x \in [0,1]}|f_n(x) - 0| = f_n\left(\frac{1}{\sqrt{2n}}\right) = \dfrac{1}{\sqrt{2e}}$. So $\displaystyle \lim_{n \to \infty} \sup_{x \in [0,1]}|f_n(x) - 0| = \dfrac{1}{\sqrt{2e}} > 0$, proving this function not converging uniformly to $0$ on $[0,1]$.

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  • $\begingroup$ How do u know the point is always positive maximum ? $\endgroup$
    – Shuzheng
    Apr 21, 2014 at 10:04
  • $\begingroup$ @NicolasLykkeIversen: because the f'(x) changes sign from + to - at critical point, and f_n at the endpoints is is less than the value of f_n at the critical value. $\endgroup$
    – DeepSea
    Apr 21, 2014 at 10:20
  • $\begingroup$ Thank you LAcarguy, very nice. $\endgroup$
    – Shuzheng
    Apr 21, 2014 at 10:24
  • $\begingroup$ For the first one, why does $\sup f_n \to 0$ imply $f_n \to 0$? How does this tell us the entire function on $[0,1]$ goes to $0$? $\endgroup$
    – user5826
    Dec 3, 2017 at 5:33
  • $\begingroup$ I think you have a small mistake when you put the value $\frac{1}{n^2}$ at the second function. It seems to be $\frac{n}{2}$ and this is not a zero sequence. So second function not uniformly convergent on $[0,1]$. $\endgroup$
    – XepheroS
    Aug 17, 2021 at 8:24

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