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?
Best Answer
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.