I need to proof that $(C_0([0,\infty)), \ \|\cdot\|_\infty)$ is complete. I think that I did it correctly (or atleast am thinking in the right direction), but I still struggle with writing it down correctly. I used that $(C_b([0,\infty)), \|\cdot\|_\infty)$ is complete, so therefore as $C_0[0,\infty) \subset C_b[0,\infty) $, I wanted to use that if $A$ is a closed subset of a complete space $(M,d)$, that $(A,d)$ is complete. Thus, it came down to showing that $C_0[0,\infty)$ is closed.
Let $f_n(k)$ be a sequence of elements out of $C_0[0,\infty)$ that converges to an element $f$ out of $C_b[0,\infty)$. I need to proof that $f(k) \in C_0[0,\infty)$.
Per the definition of convergence, we know that $\|f_n(k) - f(k) \|_\infty \rightarrow 0$ whenever $n \rightarrow \infty$. We use this to find $ |f_n(k)-f(k)| \leq\|f_n(k) - f(k)\|_\infty \rightarrow 0$.
Then we consider $|f(k)| = |f(k) - f_n(k) +f_n(k)| \leq |f(k)-f_n(k)| + |f_n(k)|$. If we take limits on both sides, we get: $\lim_{k\to \infty} |f(k)| \leq \lim_{k\to \infty} |f(k)-f_n(k)| + \lim_{k\to \infty} |f_n(k)| = 0 + 0 = 0$ (the first term follows from what we have previously stated and the second term follows from the fact that $f_n(k)$ is an element of $C_0$. Thus, $f \in C_0$ and therefore $C_0$ is closed and therefore it is complete.
My questions are the following;
What exactly does it mean to be a sequence of functions?
I used $f_n(k)$ and $f(k)$ in my proof because I saw that it was used in another exercise, but I'm not sure what it means to write that $k$ down.
Is the structure of my proof sound?
Is there a better way to proof such statements?
Thanks for reading,
K. Kamal
Suppose $f_n$ is Cauchy, then since you know $C_b$ is complete you know that $f_n \to f$ for some $f \in C_b$. All that remains is to show that $f(t) \to 0$.
Note that $|f(t)| \le |f(t)-f_n(t)| + |f_n(t)| \le \|f-f_n\| + |f_n(t)|$.
Let $\epsilon >0$ then choose $n$ such that $\|f-f_n\| < { 1\over 2} \epsilon$.
Now choose $T$ such that for $t \ge T$ we have $|f_n(t)| < { 1\over 2} \epsilon$, hence $|f(t)| \le \epsilon.$