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

Question

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!

Answer 1

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*}

Answer 2

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 $t \mapsto \frac{1}{t} \in L^2([1,+\infty))$)

Answer 3

$f \in L^2[1,\infty) \mapsto \langle f,1/t\rangle_{L^2[1,\infty)}$ 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 $M$.