1
$\begingroup$

Study the punctual and uniform convergence of $f_n(x)$ on $A$

$$f_n(x)=\frac{x}{1+n^2 x^2} \ \ \ A=[-1,1]$$

My reasoning:

Punctual convergence

$\forall x \in A $ $$ \lim_{n \to +\infty} f_n(x)=f(x) \\ f\equiv0$$

Uniform convergence

It needs this propriety: $ f_{n+1}(x) \le f_n(x) $ (Dini's theorem hypothesis)

So, the succession converges uniformly on $[0,1]$

Is this reasoning correct?

$\endgroup$
  • $\begingroup$ Yes, it is good, for the uniform convergence, quote the Dini's theorem $\endgroup$ – Tsemo Aristide Oct 18 '15 at 18:04
  • $\begingroup$ @Tsemo Aristide thanks! $\endgroup$ – Elsa Oct 18 '15 at 18:07
  • $\begingroup$ Or $\left|\frac{x}{1+n^{2}x^{2}}\right| \le \frac{1}{2n}$ for all $x \in [-1,1]$ because $|2(1)(nx)| \le 1^{2}+(nx)^{2}$. $\endgroup$ – DisintegratingByParts Oct 18 '15 at 21:32
  • $\begingroup$ @TrialAndError Thanks! So, I should consider the absolute value of $f_n(x)$, why? $\endgroup$ – Elsa Oct 19 '15 at 13:12
1
$\begingroup$

For an alternate proof, note that $f_n'(x)=0$ implies $x=\pm\frac1n$, and \begin{align} f_n\left(\frac1n\right) &= \frac1{2n}\\ f_n\left(-\frac1n\right) &= -\frac1{2n}\\ f(-1) &= -\frac1{1+n^2}\\ f(1) &= \frac1{1+n^2}. \end{align} It follows that $$\lim_{n\to\infty}\sup_{x\in[-1,1]}|f_n(x)|=0, $$ so that $f_n$ converges uniformly to $0$ on $[-1,1]$.

$\endgroup$
1
$\begingroup$

Yes, your reasoning is correct: a monotone sequence converging pointwise to a continuos function converges uniformly. In case you are interested in a proof, you can find one here. The same reasoning can be applied to prove uniform convergence in $[-1,0]$, where the sequence of functions is monotonically increasing to $0$.

Just a side note: it is not true that "it needs" that property, as you say. Indeed, you can prove prove uniform convergence directly from the definition. This can be done considering the derivative of $f_n$ and looking for where it attains the maximum distance from the limiting function. But derivatives never bothered you anyway.. :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.