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I found this problem that asks to prove that

the subspace of $l^2(\mathbb{N})$, where $M = [\{x_n\}_{n=1}^\infty∈l^2(\mathbb{N}) \mid \sum\limits_{n = 1}^\infty \frac{1+e^{in}}{n^{3/2}}x_n=0]$ is closed in $l^2(j \in \mathbb{N})$.

How should I approach this to fit the inner product definition? I would appreciate any guidance on the mechanics.

What I know:

  • $l^2$ space is closed if it contains all of its limit points.
  • Given a subspace V of a Hilbert space, $\mathcal H$, we define the orthogonal complement of V to be $V^{\perp}:=[{f\in\mathcal H |\langle f,g\rangle,∀g \in V}] $
  • $V^{\perp}$ is a closed subspace of $\mathcal H$, where $\langle f,g\rangle=0$, $∀g \in V$. Need to show that since the inner product is continuous, then $\lim_{n \to \infty}\langle f_n,g\rangle = \langle f,g\rangle$ ⇒ $\langle f,g\rangle=0$, and so $f\in V^{\perp}$.
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    Do you mean $$M = [\{x_n\}_{n=1}^\infty∈l^2|\sum\limits_{n = 1}^\infty \frac{1+e^{in}}{n^{3/2}}x_n=0]?$$ If so, try to prove that $y_n=\dfrac{1+e^{in}}{n^{3/2}}$ is in $\ell^2$ and let $V=\{y_n|n\in\mathbb{N}\}$. Then $M=V^\perp$ and so it is closed. – 2017-02-08
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    $M$ is the kernel of an obvious (continuous) linear functional. – 2017-02-08
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    For some $y \in l^2, \|y\| = 1$, let $P(x) = x - \langle x,y \rangle y$. It is an orthogonal projection on the closed subspace you want. – 2017-02-08
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    I changed $a_n$ in the sum to $x_n$. Please check to make sure this is correct, and fix it if necessary. Also, please don't use `<` and `>` for $\langle$ and $\rangle$. Use `\langle` and `\rangle` for this. That fixes the symbols and the spacing, both. – 2017-02-08

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Hint: What is the orthogonal complement of the space spanned by the vector $\bigl\{(1+e^{in})/n^{3/2}\bigr\}_{n=1}^\infty$?