Proving the metric space with integral norm is not complete

Question

Let $\text{C}([0,1])$ be the linear space of continuous functions $f:[0,1]\rightarrow\mathbb{R}$. Defined is the norm:

$||f|| := \int_0^1|f(x)|dx $.

We have a sequence of functions $(f_n)_{n\geq2}$ defined by $\frac{1}{\sqrt{x}}$ as $x>\frac{1}{n}$ and $\sqrt{n}$ as $x\leq\frac{1}{n}$ which is a Cauchy sequence.

I was asked on a test to show that $\text{C}([0,1])$ is not complete by proving there cannot be a function $g\in\text{C}([0,1])$ for which $\lim_{n\rightarrow\infty}||g-f_n||=0$. I'm not sure how to go about this.

When I write the integral like this: $\int_0^{\frac{1}{n}}|g(x)-\sqrt{n}|dx + \int_{\frac{1}{n}}^1|g(x)-\frac{1}{\sqrt{x}}|dx$ am I allowed to equate the first part to $0$ since $n$ goes to infinity?

Answer 1

This sequence tends to function $f(x) =\sqrt{\frac{1}{x}}$ on $(0,1]$ but this function cannot be extend to continuous function on whole $[0,1].$ Therefore there is not element $g\in C([0,1])$ such that $$||f_n -g||\to 0$$

Answer 2

In the first term Use the fact that $\int_0^{1/n} |g(x)-\sqrt n| \, dx \leq \int_0^{1/n} |g(x)| \, dx+ \frac 1 {\sqrt n} \to 0$ since $g$ is bounded. If $||f_n-g||$ indeed goes to $0$ the second term gives $\int_0^{1} |g(x)-\frac 1 {\sqrt x}| \, dx=0$ which implies that $g(x)= \frac 1 {\sqrt x}$ almost everywhere. However, by continuity of both sides this gives $g(x)= \frac 1 {\sqrt x}$ for all $x>0$. This makes $g$ unbounded contradicting the fact that it is continuous.