# Uniform convergence of $f_n=\frac{\sqrt{n}\cdot \sin(x)}{nx^{2}+2}$

I am learning about uniform convergence of function series. I wanted to ask whether I answered this exercise correctly:

Let $$f_n=\frac{\sqrt{n}\cdot \sin(x)}{nx^{2}+2}$$

1. Does $f_n$ uniformly converge in $[0,\infty)$
2. Does $f_n$ uniformly converge in $[-\pi,\pi]$?
3. Does $\displaystyle\int_{\pi}^{2\pi}f_n(x)dx \rightarrow 0$?

Solution

$f_n$ is point-convergent to $f(x)=0$ in R. Put $x_n=\frac{1}{\sqrt{n}}$, then $|f_n(x_n)-f(x_n)|\rightarrow \frac{1}{3}$, and therefore the answer to 1. and 2. is that $f_n$ is not uniformly convergent in those intervals.

For 3., the answer is yes since $f_n(x)\rightarrow f(x)=0$.

What is odd about this solution to me is that you can answer 1. and 2. seemingly in the same way. So, is this correct?

Thanks!

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Thank you everyone for the answers! I wish I could've chosen them all :). – ro44 Jan 27 '12 at 15:13

Your argument in (3) is wrong: Consider a sequence of piecewise linear functions $(f_n)$ where the graph of $f_n(x)$ is a straight line through $(0,0)$ and $(\frac1{2n},2n)$ for $0 \leq x \leq \frac1{2n}$ and a straight line through $(\frac1{2n},2n)$ and $(\frac1n,0)$ for $\frac1{2n} \leq x \leq \frac1n$ and $0$ everywhere else.

Then we have for all $x \in [0,1]$ that $f_n(x) = 0$ eventually. In particular $f_n \to 0$ pointwise, but: $$\lim_{n\to \infty} \int_0^1 f_n(x)\,dx = 1 \neq 0 = \int_0^1 \lim_{n\to \infty} f_n(x)\,dx.$$

This is a standard example of why pointwise convergence and integrals is a delicate question.

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Thank you for your input! I guess the answer in (3) is dependent on whether $f_n$ is uniformly convergent in $[pi,2pi]$, which I think it is since $|f_n-0|<\frac{1}{\sqrt{n}}$ in that interval. So the answer is still yes for different reasons. – ro44 Jan 27 '12 at 15:08
@ro44 Kahen's point is that you only showed pointwise convergence. It is true that if $f_n$ converges to $0$ uniformly on $[a,b]$ then $\int_a^b f_n(x)dx$ converges to $0$. In other words, your argument above is wrong, but the "yes, it does converge to zero" answer is correct. – Thomas Andrews Jan 27 '12 at 15:45

Your solution for 1) and 2) is correct. Your answer for part 3) is correct, but the reasoning is flawed. Pointwise convergence does not imply convergence of the integrals; uniform convergence, however, does. That is, if $\{ g_n\}$ converges uniformly to $g$ on $[a,b]$ and if each $g_n$ is integrable on $[a,b]$, then $g$ is integrable on $[a,b]$ and $\int_a^b g_n\rightarrow\int_a^b g$.

You can show that your $f_n$ converge uniformly to 0 on $[\pi, 2\pi]$ (which leads me to believe that part 2) should have asked if the convergence was uniform on $[\pi,2\pi]$).

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For 3. pointwise convergence isn't strong enough (as other users have already answered). The good news is that your sequence of functions converges uniformly in $[a, +\infty], \forall \ a>0$, and therefore also in $[\pi, 2\pi]$. To prove this, simply reason as follows:

$$\sup_{[a,+\infty]}{|f_n|}=\sup_{[a, +\infty]}{\frac{\sqrt{n}\cdot \sin(x)}{nx^{2}+2}} \leq \frac{\sqrt{n}}{na^{2}+2} \rightarrow 0$$

for $n \rightarrow\infty$. Your problem with uniform convergence in $\mathbb{R}$, as your solution to 1. shows, is located "close to $0$"; if you move away from $0$ this problem disappears.

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