Close $M=\{\:x(t)\in{L_2}_{[1,\infty)}:\int_{1}^{\infty}\frac{x(t)}{t}dt=0\:\}$? 
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!
 A: 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*}
A: 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 $t \mapsto \frac{1}{t} \in L^2([1,+\infty))$)
A: $f \in L^2[1,\infty) \mapsto \langle f,1/t\rangle_{L^2[1,\infty)}$ 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 $M$.
