# Is $C[0,1]$ a Banach space for the $L^1$ norm?

Is $C[0,1]$ a Banach space with respect to the norm $\|f\| = \int\limits_0^1|f(t)| \, dt$?

People keep telling me it is, but lets consider: $f_n(x) = x^n$. This function defines a Cauchy sequence, yet the limit clearly isn't a continuous function!

• $\lim f_n=0$ for this norm. Sep 30, 2016 at 11:20
• That is not a norm, only a seminorm. $\int_0^1 t - \frac{1}{2}\,dt = 0$. Should it be $\int_0^1 \lvert f(t)\rvert\,dt$? Sep 30, 2016 at 11:20
• I was under the impression it is. Anyhow, the same argument stands for the $L^1$ norm. I'll edit now. Sep 30, 2016 at 11:21

Under the $L^1$ norm, $C[0,1]$ is not a Banach space. Your example $f_n(x) = x^n$ doesn't do the trick since $\|f_n(x) - 0\| \to 0$ as $n \to \infty$, so there is a limit to this sequence in $C[0,1]$. However, it is well known that you can create a sequence $g_n \in C[0,1]$ such that $g_n \to 1_{A}$ for any interval $A \subset [0,1]$, for example $A = (1/4, 3/4)$, but there is no $g \in C[0,1]$ such that $\|g - 1_A\| = 0$.

Firstly $\Vert \cdot \Vert$ is just a seminorm (try to find an $f \in C[0,1]$ with $\Vert f \Vert = 0$ but $f \neq 0$).

The standard norm on $C[0,1]$ is given by $\Vert f \Vert_\infty = \sup_{x \in [0,1]} \vert f(x) \vert$. Respecting this norm $C[0,1]$ is a Banach space due to the Weierstrass theorem. I hope it helps you :)

• What about $f_n(x) = min(n, 1/x)$? This converges to $1/x$ which isn't a continuous function! Sep 30, 2016 at 11:28
• @yaddle $\lVert\bullet\rVert$ is a norm on $C[0,1]$. It does not make it a Banach space, but most certainly if $\lVert f\rVert =0$, then $f=0$.
– user228113
Sep 30, 2016 at 11:46
• Oh, I see: the mistake was corrected shortly before you posted the answer.
– user228113
Sep 30, 2016 at 11:48