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## Closed linear subspace

Functional Analysis
Papapetros Vaggelis
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### Closed linear subspace

For $\displaystyle{p\in\left[1,+\infty\right)}$, consider

$\displaystyle{E_{p}:=\left\{f\in L^{p}([0,+\infty))\,\,,\int_{0}^{\infty}f(x)\,\mathrm{d}x=0\right\}}$.

(Lebesgue measure)

i. Prove that $\displaystyle{E_{p}}$ is a linear subspace of $\displaystyle{L^{p}([0,+\infty))}$.

ii. Prove that $\displaystyle{E_{p}}$ is closed if, and only if, $\displaystyle{p=1}$.
r9m
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### Re: Closed linear subspace

If, $p = 1$, then $E_1$ is clearly a closed subspace of $L^1([0,\infty))$ as: If $f_n \overset{L^1}{\longrightarrow} f$ for $f_n \in E_1$, then $\displaystyle \left|\int_{0}^{\infty} f\,dx\right| = \left|\int_{0}^{\infty} (f-f_n)\,dx\right| \le \int_{0}^{\infty} \left|f - f_n\right|\,dx \underset{n \to \infty}{\longrightarrow} 0$, i.e., $f \in E_1$.

I thought about the converse for some time, there must be an argument simpler than what I have in mind.

For the converse, if $p > 1$, it suffices to show that $\displaystyle E_p^{\infty} = \left\{f \in C_c^{\infty} \cap L^p([0,\infty)): \int_0^{\infty} f\,dx = 0\right\}$ is dense in $L^p([0,\infty))$ (since, $E_p^{\infty} \subset E_p \subset L^p([0,\infty))$ are proper subspaces in that order, $L^p$-closure of $E_p$ must be $L^p([0,\infty))$).

Let, $f \in E_p$, by Lusin's theorem, $\displaystyle \exists g \in C_c^{\infty}$ such that $\lVert g - f \rVert_p < \epsilon$. If, $\displaystyle \int_0^{\infty} g \,dx = 0$ then $g \in E_p^{\infty}$ and we are done. Suppose, $\displaystyle J = \int_0^{\infty} g \,dx \neq 0$, then let, $\text{supp } g \cap [n,\infty) = \emptyset$ for some $n \ge 1$. Pick a 'standard' mollifier $\phi \in C_c^{\infty}$ supported in $[n,\infty)$ (i.e., $\text{supp } \phi \subset [n,\infty)$, $\phi \ge 0$ and $\displaystyle \int_0^{\infty} \phi \,dx = 1$) such that $\displaystyle \lVert \phi \rVert_p < \frac{\epsilon}{|J|}$.

Consider, $h = g - J\phi \in E_p^{\infty}$, which satisfies $\displaystyle \lVert f - h \rVert_p \le \lVert f - g \rVert_p + |J|\lVert \phi \rVert_p \le 2\epsilon$. This shows that $E_p^{\infty}$ is dense in $L^p([0,\infty))$.

To see the existence of such a 'standard' mollifier with required properties, using Urysohn's lemma one finds a $C^{\infty}$ function $\psi_m : (0,\infty) \to [0,1]$ such that $\psi_m\rvert_{[n+1,n+m+1]} = 1$ and $\text{supp } \psi_m \subset [n,n+m+2]$ for $n,m \ge 1$. Then $\displaystyle J = \int_0^{\infty} \psi_m\,dx \ge \int_0^\infty \chi_{[n+1,n+m+1]}\,dx = m > 0$. Then normalize, $\phi_m := \dfrac{\psi_m}{J}$, so that $\displaystyle \int_0^{\infty} \phi_m\,dx = 1$ and $\displaystyle 0 \le \phi_m \le \frac{1}{m}$. Clearly, $\displaystyle \lVert \phi_m \rVert_p \le \left[\int_n^{n+m+2} \frac{1}{m^p}\,dx\right]^{1/p} = \left(1+\frac{2}{m}\right)^{1/p}\frac{1}{m^{1-\frac{1}{p}}} \to 0$ as $m \to \infty$ (since, $p > 1$). That is $\lVert \phi_m \rVert_p$ can be made arbitrarily small by choosing large $m$.