Show that the set of continuous functions $C[a,b]$ under the metric $\rho(f,g)=\displaystyle{\sqrt{\int_a^b|f(t)-g(t)|^2dt}}$ is not a complete metric space for $f,g\in C[a,b]$ and $t\in[a,b]$.
To prove that this is incomplete, we must find a Cauchy sequence which converges to a function that is not continuous. Let $\{f_n\}$ be a Cauchy sequence. Then, we define the following:
\begin{align*}
f_n(x) = x^n\mbox{on}[0,1].
\end{align*}
Now if we take the limit as $n$ approaches infinity, we have:
$$f(x)= \lim_{n\rightarrow \infty}f_n(x) =
\begin{cases}
0,& \text{if } x\in[0,1) \\
1, & x=1
\end{cases}
$$
We know that $\{f_n\}$ is a Cauchy sequence which converges, because for any $\epsilon>0, \exists \thinspace N\in\mathbb{N}$ such that $|f_n-f_m| < \epsilon$ for all $n,m\geq N$. However, the point-wise limit seen above is not continuous, since $\{f_n\}\rightarrow f$. Thus, we have that not every Cauchy sequence converges into the set of continuous functions $C[a,b]$. We have shown that this metric space is not complete.
My problem here is I am mixing up $f_n$ with $f_n(x)$ and $f(x)$, some clarification on whether I did them right would be helpful and if there are any mistakes in the proof advice would be appreciated. Also if there are any problems with the proof please comment.
Best Answer
Looks mostly fine to me. The thing to keep in mind is that $f(x)$ or $f_n (x)$ refers to the value of the function at the point $x$, which are real numbers. By contrast, $f_n$ or $f$ by themselves represent the function as an object unto itself.
This should technically read $f_n\rightarrow f$ since the sequence approaches $f$ itself, not just a value of $f$. Similarly, your display equation should technically read $$f(x)=\lim_{n\rightarrow\infty}f_n(x)=\begin{cases} 1 &\mbox{if } x\in[0,1)\\ 0 &\mbox{if } x=1\end{cases}$$ since you are giving a pointwise definition of $f$. Hope this helps!