# Closed set in Banach?

• April 15th 2010, 01:08 AM
veljko
Closed set in Banach?
Is set
$L=\{f\in X | \int_{-\infty}^{+ \infty} f(x)dx =0\}$.
closed in Banach space $X$?

a) $X = L^1 (-\infty, +\infty)$, b) $X = L^2 (-\infty, +\infty)$
• April 15th 2010, 02:45 AM
Enrique2
a) Is true because $f\mapsto \int_{\mathbb{R}}f$ is a continuous linear functional on $L_1(\mathbb{R})$. If we denote by $T$ this continuous mapping the space $L$ is
$T^{-1}(0)$.

b) Is false. To prove it just observe that there are dense subspaces in $L_2(\mathbb{R})$ formed by integrable functions and with integral 0. Look at the Haar system Haar wavelet - Wikipedia, the free encyclopedia.

If the set of functions with integral 0 were closed, this would imply that the closure of the span of the Haar system, that is the whole $L_2(\mathbb{R})$, would be included in $L$, that certainly is not true.