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?

  • $\begingroup$ (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

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.

  • $\begingroup$ 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

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