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We've got a sequence of functions

$$f_n(x) = \frac {x}{nx^2 +4}$$

Some questions:

  1. Determine the pointwise limit of $(f_n)$. Call this function $f$
  2. Compute the maximum and minimum of each function $f_n$. Prove now that the convergence $f_n \rightarrow f$ is uniform
  3. Determine the sequence $(f'_n)$. For which $x$ does $f'_n(x)$ converge to $f'(x)$?
  4. Considering the following theorem, are the conditions satisfied? :

    Let $f_n \rightarrow f$ pointwise on the closed interval [a,b], and assume that each $f_n$ is differentiable. If $(f'_n)$ converges uniformly on [a,b] to a function g, then the function $f$ is differentiable and $f'=g$

My try

  1. $$\lim\limits_{n \to \infty}f_n(x) = \frac{x}{\infty}=0 $$
  2. Using the Quotient Rule, we have $$ f'_n(x)= \frac {nx^2+4-2nx^2}{(nx^2+4)^2} = 0$$ Looking for extrema we set $nx^2+4-2nx^2 =0 \implies -nx^2 +4 = 0 \implies x = \pm \sqrt{\frac {n}{4}}$. We have a minimum on $x=-\sqrt{\frac{n}{4}}$ and a maximum on $x=\sqrt{\frac{n}{4}}$ for all $n$. The values of the extreme are: for the minimum $$f_{n,min}=f_n(-\sqrt{\frac{n}{4}}) = \frac{-\sqrt{{\frac{n}{4}}}}{n\frac{n}{4} +4} $$ $$f_{n, max}= f_n(\sqrt{\frac{n}{4}})= \frac{\sqrt{{\frac{n}{4}}}}{n\frac{n}{4} +4}$$ So, now we know that $f_n$ is bounded between the minimum and maximum value. How can I proof that the convergence is uniform?
  3. I don't understand
  4. I have to check wheter $(f'_n)$ converges uniformly to a function g, and I have to check wheter $f$ is differentiable?
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Calculation for extremum values is wrong. The value of $x$ is $\dfrac{2}{\sqrt{n}}$. – user45841 Oct 24 '12 at 5:45
up vote 2 down vote accepted
  1. Correct.

  2. Computations are correct. To check uniform convergence you need to verify $\lim\limits_{n\to\infty}\sup\limits_{x\in\mathbb{R}}|f_n(x)-f(x)|=0$. That is why you were asked to find minimum and maximum.

  3. Compute $f'_n(x)$ and $f'(x)$, check for which $x$ holds $\lim\limits_{n\to\infty} f'_n(x)=f'(x)$

  4. Consider point $x=0$, to see that $f'_n(x)$ doesn't converges to $f'(x)$ uniformly.

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Can you explain it somewhat more elaborate? :). I'm new to this subject. For 2: Why do I have to verify that lim sup {fn(x)-f(x) = 0. For 3: What is f'(x)? 0? For 4: Can you explain what happens at x=0 ? – Applied mathematician Oct 23 '12 at 17:11
As for the paragraph 2, it is the criterion of unoform convergence. As for the 3) you are right $f'(x)=0$, because $f(x)=0$. As for the 4) I suggest you to compute $f'_n(0)$ – Norbert Oct 23 '12 at 17:17
2.) What I have to check is very easy? $f_n(x)$ = 0 as n goes to infinity, and f(x) = 0, so then we have 0 for the given limit? – Applied mathematician Oct 23 '12 at 20:23
it is not so easy as your thought becasue you need to find not the difference maximum of difference between $f_n$ and $f$ for fixed $n$. And if this difference tends to infinity then convergence is uniform – Norbert Oct 23 '12 at 20:26
$$\lim\limits_{n\to\infty}\sup\limits_{x\in\mathbb{R}}|f_n(x)-f(x)|=\lim\limits_‌​{n\to\infty}\sup\limits_{x\in\mathbb{R}}|f_n(x)-0|=$$$$\lim\limits_{n\to\infty} \sup\limits_{x\in\mathbb{R}} |f_n(x)|=$$$$=\lim\limits_{n\to\infty}\max(|f_{n,\min},|f_{n,\max}|)=\lim\limits‌​_{n\to\infty}\frac{2\sqrt{n}}{n^2+16}=0$$ – Norbert Oct 23 '12 at 20:32

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