Injectivity and surjectivity of a linear bounded operator $T : L^2(1, +\infty) \to \ell^1$

Question

I have the following exercise:

For if $f \in L^2(1,+\infty)$ consider the sequence $$ z_n := \int_{(n, n+1)} \frac{f(x)}{x} \mathrm d x \quad \text{for } n \in \mathbb N .$$ (Here, the integral is à la Lebesgue.) Then:

1. Show that the sequence $z := (z_n \mid n \in \mathbb N) \in \ell^1$.
2. In this case you have defined one function $T : L^2(1, +\infty) \to \ell^1$. Show it is linear and bounded.
3. Is $T$ injective? Is it surjective?

I have no trouble with the first two points, but I cannot figure how to conlude the last point.

For injectivity, I may consider $$\ker T = \left\{f \in L^2(1,+\infty) \colon \int_{(n, n+1)} \frac{f(x)}{x} \mathrm d x = 0 \text{ for every } n \in \mathbb N \right\}$$ If can conclude that the null function $(1, +\infty) \to \mathbb R$ is the unique one for which all these integrals are zero, then injectivity is proved. But is it true? It seems no, but I am not able to underpin this answer. As for suriectivity, I don't know what to say.

Some hints?

Answer 1

Let $f(x)={x(x-\frac 3 2)}$ for $1 \leq x <2$ and $0$ elsewhere. Then $f \in Ker (T)$ so $T$ is not injective.

For any $f\in L^{2}$ we have $|z_n|^{2} \leq \int_n^{n+1} |f(x)|^{2}dx \int _n^{n+1} \frac 1 {x^{2}}dx$ which gives $|z_n| =o(\frac 1{n})$. So any sequence in $\ell^{1}$ which does not satisfy this inequality is not in the range of $T$. So $T$ is not surjective.