The Weierstrass $M$-test says that if $(f_n$) is a sequence of functions on $E$, and $\left|f_n(x)\right|\leq M_n$, then $\sum f_n$ converges uniformly on $E$ if $\sum M_n$ converges.

My question is, can $M_n$ depend on $x$? that is, if I know $\left|f_n(x)\right|\leq \dfrac{1}{x^2n^2}$ (whose sum does converge), is this sufficient?


For any $\;x\in[a,b]\;$ , we have that

$$\left|\frac x{1+n^2x^2}\right|=\frac{|x|}{1+n^2x^2}\le\frac{|b|}{1+n^2\min\{|a|,|b|\}^2}$$

and the series


converges as long as that minimum is not zero, so we have uniform convergence in such bounded intervals which don't include zero.

Thinking a second time, if $\;x=0\;$ then the function is zero so everything's fine, too.

Added: For $\;x\ge 0\;$ we have

$$f_n(x):=\frac x{1+n^2x^2}\implies f'_n(x)=\frac{1-n^2x^2}{(1+n^2x^2)^2}=\frac{(1-nx)(1+nx)}{(1+n^2x^2)^2}=0\iff x_0=\frac1n$$

and clearly we have for values of $\;x\;$ close to $\;x_0\;$


from which we get that $\;x=x_0=\frac1n\;$ is a maximal point of $\;f_n(x)\;$ on $\;[0,\infty)\;$ , and since $\;f_n(-x)=-f_n(x)\;$ the same is true in $\;(-\infty,0)\;$, and from here we get that for all $\;x\in\Bbb R\;$

$$\left|f_n(x)\right|= \left|\frac x{1+n^2x^2}\right|\le \left|f_n\left(\frac1n\right)\right|=\frac1{2n}\xrightarrow[n\to\infty]{}0\implies f_n(x)\stackrel{\text{unif. on}\;\Bbb R}{\xrightarrow[n\to\infty]{}}f(x)=0$$

  • $\begingroup$ I'm trying to show this for the interval $[0,1]$. So if $x=0$ then as you say, everything's fine. If not, then $x\in [a,1]$ for some $a>0$. So we can make $M_n=\dfrac{1}{1+n^2a^2}$, which converges. What I fail to see is, how does letting this depend on $a$ differ from letting it depend on $x$? $\endgroup$ – user5675443 Feb 9 '15 at 2:16
  • $\begingroup$ What the above shows is uniform convergence in any closed interval. I think it can be shown in fact u.c. in the whole real line. Perhaps later I'll add something about this. $\endgroup$ – Timbuc Feb 9 '15 at 2:20
  • $\begingroup$ @user5675443 you can read now the stuff I added. $\endgroup$ – Timbuc Feb 9 '15 at 9:40

No, $M_n$ cannot depend on $x$. That's the whole point of the M-test. In your case, you should fix a range for $x$ to obtain a bound that is independent on it.

In other words: If you allow $x$ to be very small your bound becomes meaningless.

  • $\begingroup$ That's what I thought, thanks. Any ideas on how I can bound $\dfrac{x}{1+x^2n^2}$ not depending on $x$, in order to apply the test? That is, if I can; I am not sure if $\sum\limits_{n=1}^\infty \dfrac{x}{1+x^2n^2}$ converges uniformly. $\endgroup$ – user5675443 Feb 8 '15 at 21:18
  • $\begingroup$ @user5675443 You should add your last comment's question in the body of your main post. $\endgroup$ – Timbuc Feb 8 '15 at 21:34

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