Problem: $f_n(x) = \frac{6}{(1+x^{2n})}$ and $x\in\mathbb{R}$.

Find all real numbers $x$ where $f_n(x)$ converges and describe the limit function.

I found the limit function to be $$f(x)=\begin{cases}0&\text{ if }|x|>1,\\ 6&\text{ if }0 \leq |x| < 1,\\ 3&\text{ if }|x| = 1.\end{cases}$$

So $f_n(x)$ converges for all $x\in\mathbb{R}$.

Now I have to find the intervals for which $f_n$ converges uniformly to $f$. I am considering the intervals $[0,1)$ and $(1,\infty)$. But I do not know where to go from here. I am thinking the M-test? But I am not sure how to apply it. Any suggestions or hints would be helpful.

  • $\begingroup$ Hint: Draw the graph of $f$. If $f_n$ converges uniformly to $f$ in a certain interval, then $f$ is continuous in that interval. $\endgroup$ – Luiz Cordeiro May 4 '16 at 0:37
  • $\begingroup$ when I draw the graph, f is continuous on [0,1) and (1,infinity)... why does this mean fn converges uniformly on these intervals? $\endgroup$ – Harry May 4 '16 at 0:42
  • $\begingroup$ It does not mean that $f_n$ converges uniformly on these intervals (in fact, it doesn't). However it gives you some suggestions of which intervals does $f_n$ converge uniformly on. Try to prove that $f_n$ converges uniformly on intervals $[-\epsilon,\epsilon]$ for $0<\epsilon<1$, but $f_n$ does not converge uniformly on $(-1,1)$ (this does not finish the problem, since there are other intervals on which $f_n$ converges uniformly). $\endgroup$ – Luiz Cordeiro May 4 '16 at 0:48

Enough to consider $x\ge0$.

Next, instead of comparing with limit function, we look at increments:

$$|f_n (x)- f_m(x)| = \left|\frac{6}{1+x^{2n} } - \frac{6}{1+x^{2m}}\right| =6 \frac{|x^{2m} - x^{2n}|}{(1+x^{2m})(1+x^{2n})}. $$

Recall that the convergence is uniform on a set $A$ if and only if $$\lim_{m\to\infty} \sup_{n \ge m} \sup_{x\in A} |f_n (x)-f_m(x)|=0.$$

Without loss of generality, assume $n>m$.

1a. Let $a>1$. On $[a,\infty)$ we have uniform convergence because RHS is bounded above by $6\frac{ x^{2n}}{1+x^{2n}} \frac{6}{a^{2m}}\le 6a^{-2m}$.

1b. On $(1,\infty)$ we don't have uniform convergence because for example, looking at $x=2^{1/m^2}$, and $n=m^2$, the righthand side becomes

$$ 6 \frac{|2^{2/m} - 2^{1}|}{(1+2^{2/m})(1+2^{1})}\underset{m\to\infty} {\to} 6 \frac{ 2-1}{2(1+2)}=1.$$

2a. Let $a<1$. On the interval $[0,a]$ we again have uniform convergence. This is easy to see because righthand side is bounded above by $6a^{2m}$.

2b. On the interval $[0,1)$ there is no uniform convergence: similarly to 1b, take $x=2^{-1/m^2}$, $n=m^2$ to obtain

$$6 \frac{|2^{-2/m}-2^{-1}|}{(1+2^{-2/m})(1+2^{-1})}\underset{m\to\infty} {\to}6\frac{\frac12}{2 \times \frac 32}=1.$$

Also the sequence trivially converges uniformly on the degenerate interval $[1,1]$.


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