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Show that $M$ is closed in ${L_2}_{[1,\infty)}$

$$M=\left\{\:x(t)\in{L_2}_{[1,\infty)}:\int_{1}^{\infty}\frac{x(t)}{t}dt=0\:\right\}$$

I thought of using the Arzela-Ascoli theorem to prove $M$ is compact then conclude it is closed. However I have no idea on how to address equicountinuity in a set like $M$. To prove the function is equincontinuous:

$$\delta>0\:,\:\epsilon>0,\qquad|t-t_0|<\delta\implies\|x(t)-X(t_0)\|_{{L_2}_{[1,\infty)}}<\epsilon\:\:\:\forall x(t)\in M$$

Question:

1) How should I prove equicountinuity on this case?

2) Are there alternative methods? What are those?

Thanks in advance!

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Pedro Gomes
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    I am maybe mistaken but it seems to me that Arzela-Ascoli only gives relative compacity and in particular does not say that $M$ is closed. – Delta-u Apr 24 '18 at 20:45

3 Answers3

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Assume that $x_{n}\rightarrow x$ in $L^{2}[1,\infty)$ which $x_{n}\in M$, then \begin{align*} \left|\int_{1}^{\infty}\dfrac{x(t)}{t}dt\right|&=\left|\int_{1}^{\infty}\dfrac{x(t)-x_{n}(t)}{t}dt+\int_{1}^{\infty}\dfrac{x_{n}(t)}{t}dt\right|\\ &=\left|\int_{1}^{\infty}\dfrac{x(t)-x_{n}(t)}{t}dt\right|\\ &\leq\left(\int_{1}^{\infty}\dfrac{1}{t^{2}}dt\right)^{1/2}\|x_{n}-x\|_{L^{2}[1,\infty)}\\ &\rightarrow 0. \end{align*}

user284331
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  • So you prove that $x_n\to x$ defines a convergent sequence in $M$, therefore $M$ is a complete subspace of complete linear space hence it is closed. Am I right? – Pedro Gomes Apr 24 '18 at 21:02
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    Okay, you can view it like that in this case. By the way, there is a sequential characterization of closedness in normed space. – user284331 Apr 24 '18 at 21:03
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Let: $$\phi:L^2 \to \Bbb R, f \mapsto \int_1^\infty \frac{f(t)}{t}$$ then $\phi$ is a linear form and $M=\ker(\phi)$, so the closure of $M$ is equivalent to the continuity of $\phi$.

And by Cauchy-Swchartz: $$|\phi(f)| \leq \int_1^\infty \frac{1}{t} |f(t)| dt \leq \sqrt{ \int_1^\infty \frac{1}{t^2} dt}\sqrt{ \int_1^\infty |f(t)|^2 dt}=1 \times \|f\|_{L^2}$$ so $\phi$ is continuous.

(In fact $\phi(f)=\left\langle t \mapsto \frac{1}{t} , f \right\rangle$ and $t \mapsto \frac{1}{t} \in L^2([1,+\infty))$)

Delta-u
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  • Please explain me how the closure is related to the continuity. I am missing that point. – Pedro Gomes Apr 24 '18 at 21:01
  • A linear form is continuous iff its kernel is closed (see this post ).

    Here you need only the "easy" direction: if $\phi$ is continuous then: $$\phi^{-1} ( {0 } )$$ is closed as the inverse of a closed set (${0 }$) by a continuous function ($\phi$).

     – Delta-u Apr 24 '18 at 21:07
  • Are you aware that the spaces are infinite dimensional? – Pedro Gomes Apr 25 '18 at 12:38
  • Yes, but it does not change the fact that the inverse of a closed set by a continuous function is a closed set or the fact that a linear $\phi$ form is continuous iff $|\phi(f)| \leq C | f|$. – Delta-u Apr 25 '18 at 13:00
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$f \in L^2[1,\infty) \mapsto \langle f,1/t\rangle_{L^2[1,\infty)}$ is a continuous function, and you're looking at the inverse image of the closed scalar set $\{ 0\}$ under this continuous linear functional. So, this inverse image is closed, and that's your set $M$.

Disintegrating By Parts
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