Understanding why the space of $L^2$-bounded continuous functions on $\mathbb R^n$ is incomplete So I have just started a course which threw us right into the space that contains all continuous real valued functions. In other words, for $1 \leq p <\infty$, consider
$$\left\{ f : \mathbb{R}^n \rightarrow \mathbb{C} \, : \, f \text{ is continuous, }\int_{R^n} |f(x)|^p \, dx < \infty \right\} $$
Now, I want to see why this space is incomplete. Here is a piece of work that I am reading and trying to understand. Could someone break it down for me?

There is a couple of things I don't understand. One is where is the upper bound $4 \times \frac{2}{k}$ coming from? Also what is the limit of this seq? How do I know it doesn't exist in the space itself?
Thanks
 A: The easiest way to see that it is not complete is to notice that if $f_n\rightarrow f$ in $\|\cdot\|_p$ then $f_n(x)\rightarrow f(x)$ for almost all $x\in\mathbb{R}$. So essentially any $\|\cdot\|_p$ limit must also be a pointwise limit.
Now if you look at your example it  is not too hard to see that $f^{(k)}$ converges pointwise to a function that is not continuous, therefore any $\|\cdot\|_p$ limit point of the sequence must be equal to this non-continuous function, and thus it is not in $A$.
I hope this helps!
A: Let $C = \{ f | f \text{ is continuous},\, \int |f|^2 < \infty \}$ with the norm $\|f\| = \sqrt{\int |f|^2}$.
Let $f_k(x) = \begin{cases} \operatorname{sgn} x, & |x|> \frac{1}{k} \\
kx, & \text{otherwise} \end{cases}$. Then it should be clear that $f_k \in C$ for all $k$.
Now suppose $n>m\ge N$ and evaluate $\|f_n-f_m \|^2 $ as follows:
\begin{eqnarray}
\|f_n-f_m \|^2 &=& \int |f_n-f_m|^2 \\
&=& \int_{-\infty}^\infty |f_n(x)-f_m(x)|^2 dx \\
&=& 2 \int_{0}^\infty |f_n(x)-f_m(x)|^2 dx \\
&=& 2 \int_{0}^{\frac{1}{m}} |f_n(x)-f_m(x)|^2 dx \\
&=& 2 ( \int_{0}^{\frac{1}{n}} |f_n(x)-f_m(x)|^2 dx + \int_{\frac{1}{n}}^{\frac{1}{m}} |f_n(x)-f_m(x)|^2 dx) \\
&=& 2 ( \int_{0}^{\frac{1}{n}} (nx-mx)^2 dx + \int_{\frac{1}{n}}^{\frac{1}{m}} (1-mx)^2 dx) \\
&=& 2(\frac{(n-m)^2}{3 n^3} + \frac{(n-m)^3}{3 m n^3}) \\
& \le & 2 ( \frac{n^2}{3 n^3} + \frac{n^3}{3 m n^3} ) \\
&=& \frac{2}{3}(\frac{1}{n}+ \frac{1}{m}) \\
&\le & \frac{4}{3} \frac{1}{N}
\end{eqnarray}
from which it follows that the sequence $f_n$ is Cauchy.
Proving that $f_n$ does not converge to a member of $C$ is a bit tedious. We proceed by contradiction and suppose the sequence converges to some $f \in C$.
The idea is to show that for $x \neq 0$, we must have $f(x) = \operatorname{sgn} x$, which contradicts $f \in C$.
Choose $x > 0$, and suppose $f(x) \neq 1$. Since $f$ is continuous, we have some $\epsilon>0$ and some $\delta >0$ such that $|f(y)-1| \ge \epsilon$ for  all $y \in (x-\delta, x+\delta)$. (We can assume that $0 < x-\delta$.)
For $n$ sufficiently large, we have $f_n(y) =1$ for all $y \in (x-\delta, x+\delta)$. Then we have the estimate $ \|f_n-f\|^2 \ge \int_{(x-\delta, x+\delta)} |f(y)-1|^2 dy \ge 2 \epsilon^2 \delta >0$, which contradicts the convergence of $f_n$ to $f$. Hence $f(x) = 1$. Similarly we have $f(x) = -1$ for $x <0$. However, this contradicts $f \in C$, hence $f_n$ cannot converge to an element of $C$.
