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!
[Math] Is $C[0,1]$ a Banach space for the $L^1$ norm
banach-spacesfunctional-analysis
Best Answer
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 $\|g - 1_A\| = 0$.