1
$\begingroup$

Suppose that $\{f_n(x)\}_{n=1}^\infty$ is a sequence of non-negative continuous real-valued functions on $\mathbb R$ such that $f_{n+1}(x) \le f_n(x)$ for all $n \ge 1$ and all $x\in \mathbb R$.

a) Prove that there is a unique function $f(x)$ on $\mathbb R$ such that $\lim f_n(x) = f(x)$ for all $x \in \mathbb R$.

b) Must $f_n \to f$ uniformly on the closed interval $[0, 1]$?

a.) Since $f_{n+1}(x) \le f_n(x)$ for all $x \in \mathbb R$, and each $f_n(x) \ge 0$, then the sequence $\{f_n(x)\}_{n=1}^\infty$ is bounded below by the zero function.

Since $\{f_n(x)\}_{n=1}^\infty$ is bounded below, it converges and since limits are unique, there exists $f(x)$ such that $\lim f_n(x) = f(x)$.

b.) If $f$ is not continuous and each $f_n$ is continuous, then the sequence will not converge uniformly.


I am not sure about my proof of part (a). Is it correct?

$\endgroup$
2
  • $\begingroup$ (b) is Dini's theorem. $\endgroup$ Aug 22 '18 at 18:38
  • $\begingroup$ In the part a, you need to explicitly state that $x$ is arbitrary but fix. $\endgroup$
    – Our
    Aug 22 '18 at 18:43
2
$\begingroup$

a) You didn't express yourself very well. The sequence is bound bellow because it is assumed that all functions are non-negative. Then, you use the fact that, for each $x$; $\bigl(f_n(x)\bigr)_{n\in\mathbb N}$ is monotonic and decreasing to prove that the limit $\lim_{n\to\infty}f_n(x)$ exists.

b) You should provide a concrete example so that your answer is complete. Take $f_n(x)=\left(\frac 1{1+x^2}\right)^n$, for instance.

$\endgroup$
6
  • $\begingroup$ Dini's theorem says the sequence of functions must converge pointwise to a continuous function. What if the limit $f(x)$ is not continuous? $\endgroup$ Aug 22 '18 at 18:44
  • $\begingroup$ like the sequence of functions $f_n(x)=x^n$. $\endgroup$ Aug 22 '18 at 18:45
  • $\begingroup$ @AlJebr You are right. I shall edit my answer. $\endgroup$ Aug 22 '18 at 18:45
  • $\begingroup$ For part (a), I should write: let $x_0 \in \mathbb R$ be arbitrary. Then since $f_{n+1}(x_0) \le f_n(x_0)$, and $f_n(x_0) \ge 0$, the sequence $\{f_n(x_0)\}$ is bounded below and hence converges by monotonocity. Hence $\{f_n(x)\}$ converges pointwise. $\endgroup$ Aug 22 '18 at 18:47
  • $\begingroup$ Is this sufficient? $\endgroup$ Aug 22 '18 at 18:48

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.