Direct sum of Hilbert Space $\ell^{2}$

Question

Let $\ell^{2}$ be the Hilbert space over $\mathbb{C}$. With inner product \begin{equation} (x,y) = \sum_{k=1}^{\infty} x_{k}\bar{y_{k}} \end{equation} Consider the mapping $f:\ell^{2} \to \mathbb{C}$ given by \begin{equation} f(x) = \sum_{k=1}^{\infty} \frac{x_{k}}{k},\qquad (x = (x_{k})_{k\in \mathbb{N}} \in \ell^{2}) \end{equation} Now let $\mathcal{N}(f)$ be the nullspace of $f$ and let $\langle v \rangle$ be the span of a vector $v \in \ell^{2}$. Show there exists a vector $y \in \ell^{2}$ such that $\mathcal{N}(f) + \langle y \rangle = \ell^{2}$

Answer 1

$f:\ell^{2} \to \mathbb{C}$ is linear and there is $y \in \ell^2$ such that $f(y) \ne 0$, hence $f( \ell^2)= \mathbb C$.

The quotient space $ \ell^2/\mathcal{N}(f)$ is isomorphic to $f( \ell^2)= \mathbb C$.

Therefore $ \dim \ell^2/\mathcal{N}(f)=1$.

This gives

$$\mathcal{N}(f) \oplus \langle y \rangle= \ell^2.$$