Is $M = \{ f \in L_2[1,\infty) : \int_1^{\infty}\frac{f(t)}{t}dt = 0 \}$ a closed subspace in $L_2[1,\infty)$?

Question

My attempt was rather too easy:

We can denote $g(t) = \frac{1}{t}$ and rewrite $M = \{ f \in L_2[1,\infty) : \langle f, \bar{g} \rangle = 0 \}$

Let $(f_n)_n \subset M$ such that $f_n \rightarrow f$ then by continuity of inner product we get $0 = \langle f_n,\bar{g} \rangle \rightarrow \langle f, \bar{g} \rangle \in M$

I feel like this is too easy to be true, mainly because the proof isn't dependent of $g$, though not sure where I am wrong. How should we prove\disprove it?

Answer 1

Your proof is fine: your proved that if $g$ is in $L_2[1,\infty)$, then $M_g = \{ f \in L_2[1,\infty) : \langle f, \bar{g} \rangle = 0 \}$ is a closed subspace of $L_2[1,\infty)$. Alternatively, this is also the kernel of a linear continuous functional, namely, $f\mapsto \langle f, \bar{g} \rangle$.