Show that $C([0, 1])$ equipped with $\Vert f\Vert_1:=\int_0^1|f(x)|dx$ is not complete

Question

Show that the normed space of continuous real valued functions $C[0,1]$ equipped with the norm $\Vert f\Vert_1:=\int_0^1|f(x)|dx$ is not complete.

---

Let be $f_n:[0,1]\to\mathbb{R}$ with $f_n(x)=x^n$. It is obvious that $f_n\in C[0,1]$ for all $ n\in\mathbb{N}$. If we consider $n,m\in \mathbb{N}$, then

$$ \Vert f_n-f_m\Vert_1=\int\limits_0^1|x^n-x^m|~dx\leq \int\limits_0^1x^n~dx+\int\limits_0^1x^m~dx=\frac{1}{n+1}+\frac{1}{m+1}, $$ which me can make arbitraryly small. This shows that $(f_n)_{n\in\mathbb{N}}$ is a Cauchy sequence. Now, we define $f:[0,1]\to\mathbb{R}$, where $$ f(x)=\begin{cases}0,&x\in[0,1)\\1,&x=1,\end{cases} $$ which is clearly a discontinuous function. However, $$ \Vert f_n-f\Vert_1=\int\limits_0^1|x^n-f(x)|~dx\leq \int\limits_0^1x^n~dx+\int\limits_0^1f(x)~dx=\frac{1}{n+1}+0, $$ which me can make arbitraryly small. So $\lim\limits_{n\to\infty}(f_n)_{n\in\mathbb{N}}= f(x)$. As $f(x)\notin C[0,1]$ the normed space $(C[0,1],\Vert f\Vert_1)$ is not complete.

---

I know that there already exist a lot of questions about this particular problem but I was wondering if my counter example works? The counter examples I have found so far seem unnecessary complicated to me; sometimes they fill various pages... So I am wondering if I am missing something? Maybe my counterexample isn't correct?

Answer 1

No, your example does not work. Because your sequence converges to $0$ in the $1$-norm.

The problem is that the $1$-norm cannot see what happens at a single point, it sees your $f$ and $0$ as the same function. That's why you need an example where the Cauchy sequence converges to a function that is discontinuous without an avoidable discontinuity; something like $f=1_{[0,\frac12]}$.

Answer 2

@Martin Argerami is right. Indeed, $\|f_n\|_1 = \frac{1}{n + 1} \rightarrow 0$. The simple convergence doesn't matter and doesn't mean any thin in this context. You indeed have $f_n \rightarrow 0$ for the $L^1$ norm despite we don't have simple convergence.

If you want a counter example, consider for example $f_n$ the affine by part function such that $f_n(x) = 0$ when $x \leqslant \frac{1}{2} - 2^{-n}$, $f_n(x) = 1$ when $x \geqslant \frac{1}{2} + 2^{-n}$ and $f_n$ is affine between on $I_n = \left]\frac{1}{2} - 2^{-n},\frac{1}{2} + 2^{-n}\right[$. If $x > \frac{1}{2}$, $f_n(x) \rightarrow 1$ and if $x < \frac{1}{2}$, $f_n(x) \rightarrow 0$ but one more, simple convergence is meaningless. Let us show that $(f_n)$ is Cauchy and diverges.

First of all, when $N$ is a large integer and $n,m \geqslant N$, we have $f_n = f_m$ outside $I_N$ and both $f_n$ and $f_m$ are bounded by $1$ thus, $$ \|f_n - f_m\|_1 = \int_0^1 |f_n(x) - f_m(x)| \, dx = \int_{I_N} |f_n(x) - f_m(x)| \, dx \leqslant \int_{I_N} 2 \, dx = 2^{2 - N} \rightarrow 0. $$ This proves that $(f_n)$ is Cauchy. Now, let us prove it diverges, and as we saw, it is not enough to prove simple convergence toward a discontinuous function, at least when the limit has integral $0$. Assume that $f_n \rightarrow f$ for some $f$.

Let $\varepsilon > 0$. We have, for $n$ large enough $2^{-n} < \varepsilon$ so $f_n = 0$ on $\left[0,\frac{1}{2} - \varepsilon\right[$, $$ \int_0^{\frac{1}{2} - \varepsilon} |f(x)| \, dx = \int_0^{\frac{1}{2} - \varepsilon} |f_n(x) - f(x)| \, dx \leqslant \int_0^1 |f_n(x) - f(x)| \, dx \rightarrow 0. $$ Therfore, $\displaystyle \int_0^{\frac{1}{2} - \varepsilon} |f(x)| \, dx = 0$ and because $f$ is continuous (it is not true otherwise !), it implies $f = 0$ between $0$ and $\frac{1}{2} - \varepsilon$. When $\varepsilon \rightarrow 0$, we deduce that $f\left(\frac{1}{2}\right) = 0$, still by continuity of $f$. Symmetrically, we can prove that $f = 1$ on any interval on the form $\left[\frac{1}{2} + \varepsilon,1\right]$ thus $f\left(\frac{1}{2}\right) = 1$, which is a contradiction. $(f_n)$ diverges.