Prove that a Cauchy sequence of piecewise functions does not converge unde the $L_1$ norm. I want to prove that the following sequence of functions does not converge and this will prove that the space of continous functions $(C([0,1]),||\cdot||_1)$ is not complete, so the sequence is:
$$\displaystyle f_{n_{1}}(x) = \begin{cases}\sqrt n & 0\le x<\dfrac{1}{n}\\\dfrac{1}{\sqrt x} & \dfrac{1}{n}\le x\le 1.\end{cases}$$
I have already prove that it is Cauchy, but How can I prove that it does not converge?, the same problem happens with the following function:
$$\displaystyle f_n(x) = \begin{cases} 1 & x\geq1/2\\0 & x\le\dfrac{1}{2}-\dfrac{1}{n} & \\ 
n(x-\dfrac{1}{2}+\dfrac{1}{n}) &\dfrac{1}{2}-\dfrac{1}{n} \leq x \leq \dfrac{1}{2}\end{cases}$$
I don't know how can I prove that this Cauchy under the $L^{2}$ norm and is not convergent this is to prove that $(C([0,1]),||\cdot||_2)$ is not complete
Can you help me with this please (both functions)?
NOTE: I can't use that $\{f_{n_{1}}\}$ converges to $1/ \sqrt(x)$
Thanks a lot in advance.
 A: For the first problem: Suppose $f_n \to f\in C([0,1])$ in this norm. Suppose $f(x_0)\ne 1/\sqrt x_0$ for some $x_0\in (0,1).$ Because $f$ is continuous at $x_0,$ we have $f \ne 1/\sqrt x$ in some interval $[x_0,b],$ where $x_0<b<1.$ Thus
$$\tag 1 \int_{x_0}^b|1/\sqrt x - f(x)|\,dx > 0.$$
Now if $n > 1/x_0,$ then $f_n(x) = 1/\sqrt x, x\ge x_0.$ Thus $\|f_n-f\|_1$ is at least as large as $(1)$ for all such $n.$ That's a contradiction. We conclude $f(x) = 1/\sqrt x$ on $(0,1).$ Therefore $f$ is unbounded, hence $f\not \in C([0,1],$ contradiction.

Second problem sketch: To show $f_n$ is Cauchy, note that if $m<n,$ then
$$\int_0^1|f_n-f_m|^2 = \int_{1/2-1/m}^{1/2}|f_n-f_m|^2 \le 1\cdot(1/m).$$
Suppose $f_n \to f\in C([0,1])$ in this norm. An idea close to that for the first problem shows $f= 0$ on $[0,1/2), f=1$ on $[1/2,1],$ contradiction.
A: You can show the second sequence is $L^2$-Cauchy by a direct calculation, as $$||f_n-f_{n+1}||^2_2= \int_{a_n}^{a_{n+1}}f_n(x)^2 dx+ \int_{a_{n+1}}^{1/2}(f_n(x)-f_{n+1}(x))^2 dx = O(n^{-3})$$   where $$a_n=1/2-1/n$$  .  So $$||f_n-f_{n+1}||_2 =  O({n^{-3/2}})$$ which suffices....................... For the first sequence, if $g \in C[0,1]$ then $g(x) \ne 1/ \sqrt x$ for some $x \in (0,1]$  otherwise $g$ is discontinuous at zero .Now the continuity of $g$ implies that, for some $k>0$  and some $y \in (0,x)$, we have $|g(z)-1/ \sqrt z | >k$ for all  $z \in [y,x]$. This implies $$||g-f_n||_1 > k(x-y)$$ for all but finitely many $n$, so $g$ cannot be the  $L^ 1$ limit of $(f_n)_{n \in N}$.................... You may use this method for the second sequence :   If $g \in C[0,1]$ there exists $x \in [0,1/2)$ with $g(x) \ne 0 $,or some $x \in (1/2,1]$ with $g(x) \ne 1$   Now show that $g$ cannot be the $L^2$ limit of $(f_n)_{n \in N}$
