Prove some facts about the space of continuous functions in the interval $[0,1]$ I want to prove the following:
Let  be the sequence of functions in $C([0,1])$ given by
$$\displaystyle f_n(x) = \begin{cases}\sqrt n & 0\le x<\dfrac{1}{n}\\\dfrac{1}{\sqrt x} & \dfrac{1}{n}\le x\le 1.\end{cases}$$
Then $f_n$ is a Cauchy sequence in $(C([0,1]),||\cdot||_1)$ that does not converge. Thus, $(C([0,1]),||\cdot||_1)$ is not complete.
So to prove that it is Cauchy I did the following:
$$\int_{0}^{1}|f_{n}(x)-f_{m}(x)|dx \leq \int_{0}^{1}|f_n(x)|+\int_{0}^{1}|f_m(x)|=\int_{0}^{1/n}|\sqrt(n)|+\int_{1/n}^{1}|1/ \sqrt(x)|+\int_{0}^{1/m}|\sqrt(m)|+\int_{1/m}^{1}|1/ \sqrt(x)|=1/\sqrt(n)+1/\sqrt(m)+(2-2/\sqrt(n))+(2-1/\sqrt(m))=4-(1/\sqrt(n)+1/\sqrt(m))$$
So the thing is, I think there is something wrong here since I can't bound $1/\sqrt(n)+1/\sqrt(m)$ as both $n,m$ goes to zero, and the norm could be negative since $1/\sqrt(n)+1/\sqrt(m)$ could be larger that 4.
Now another thing, How can I prove that is not convergent?. 
And finally, I was trying to figure out if $(C([0,1]),||\cdot||_2)$ is complete? but the thing is that proving it should be worse that rearrange the above function to give a counter example, but Am I right?.
Can someone help with this questions please?  
 A: For simplicity, let $n \wedge m := \min \{n,m\}$ and let $n \vee m := \max \{ n,m \}$ for all integers $n,m \geq 1$. 
We may do it softly. Note that
$$
\int_{0}^{1} |f_{n} - f_{m}| = \int_{0}^{\frac{1}{n \vee m}} | \sqrt{n} - \sqrt{m}| dx + \int_{\frac{1}{n \vee m}}^{\frac{1}{n \wedge m}} | \frac{1}{\sqrt{x}} - \sqrt{n \wedge m} | dx \leq \frac{\sqrt{n}}{n \vee m} + \frac{\sqrt{m}}{n \vee m} + 2\frac{1}{n \wedge m} + 2\frac{1}{n \vee m} + \frac{\sqrt{n \wedge m}}{n \wedge m} + \frac{\sqrt{n \wedge m}}{n \vee m} \to 0
$$
as $m,n$ grows.
A: Let $\varepsilon > 0$, and choose $N >\frac{1}{\varepsilon^2}$. Then for $n > m > N$, we have
\begin{align*}
\int_0^1 |f_n(x) - f_m(x)| \, \mathrm{d}x &= \int_0^1 f_n(x) - f_m(x) \, \mathrm{d}x \\
&= \int_0^1 f_n(x) \, \mathrm{d}x - \int_0^1 f_m(x) \, \mathrm{d}x \\
&= \left(\int_0^{1/n} \sqrt{n} \, \mathrm{d}x + \int_{1/n}^1 \frac{1}{\sqrt{x}} \, \mathrm{d}x\right) - \left(\int_0^{1/n} \sqrt{n} \, \mathrm{d}x + \int_{1/n}^1 \frac{1}{\sqrt{x}} \, \mathrm{d}x \right) \\
&= \left(\frac{\sqrt{n}}{n} + 2 - \frac{2}{\sqrt{n}} \right) - \left (\frac{\sqrt{m}}{m} + 2 - \frac{2}{\sqrt{m}}   \right ) \\
&= \frac{1}{\sqrt n} - \frac{2}{\sqrt{n}} - \frac{1}{\sqrt m} + \frac{2}{\sqrt{m}} \\
&= \frac{1}{\sqrt m} - \frac{1}{\sqrt n} \\
&< \frac{1}{\sqrt m} < \varepsilon,
\end{align*}
showing the sequence is Cauchy w.r.t. $||\cdot||_1.$ Now suppose $f_n$ has limit $f \in C([0,1])$. Then $f(x) = \frac{1}{\sqrt x}$ on $(0,1)$, which cannot be extended continuously onto $[0,1]$. Thus the space is incomplete.
