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 $\mathcal{N}(f) + \langle y \rangle = \ell^{2}$
Direct sum of Hilbert Space $\ell^{2}$
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hilbert-spaces
lp-spaces
direct-sum
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1What have you tried so far? Do you know of any "easier" versions of this (potentially for a finite dimension inner product space)? – 2017-01-18
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0In general I know that if we have a closed subspace of a Hilbert space we can write the hilbert space as a direct sum of the closed subspace and its orthogonal complement. The nullspace of f is closed (which can easily be shown). After that I was trying to show that the span of y equals the orthogonal complement of the null space of f. However, I didn't manage in doing that. – 2017-01-18
1 Answers
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$f:\ell^{2} \to \mathbb{C}$ 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
$$\mathcal{N}(f) \oplus \langle y \rangle= \ell^2.$$