Consider the inner product space $l_0$ consisting of all infinite sequences of complex numbers with only finitely many non-zero terms, with the inner product of $l^2$ (space of square summable sequences): $$<x,y>=x_1\overline y_1 +x_2\overline y_2 + ...$$ for $x = (x_1,x_2,...,x_n,...)$ and $y =(y_1,y_2,...,y_n,...)$.
Let $a=(1,\dfrac12,\dfrac1{2^2},\dfrac1{2^3},...)\in l^2$.
$1)$ Show that $W = \{x\in l_0 : <x,a>=0\}$ is a subspace of $l_0$. Is $W$ finite dimensional or infinite dimensional?
$2)$ Show that $W^\perp=\{y\in l_0: <x,y>=0 \text{ for all } x\in W\}=\{0=(0,0,...0,...)\}$.
I have shown that $W$ is a subspace of $l_0$
And I can find a linearly independent set of sequences $\{b_1,b_2,...\} \subseteq W$ which spans the space of all sequences.
$b_1=(1,-2,0,0,.....)$
$b_2=(0,1,-2,0,0,....)$
$b_3=(0,0,1,-2,0,0,...)$
...
$b_n=(0,0,0,...,1,-2,0,0,...)$
Any sequence $(x_1,x_2,x_3,...)$ can be expressed as linear combination of $\{b_1,b_2,...\}$ hence they also span $l_0$.
Is this enough to conclude that $W$ is infinite dimensional since it is spanned by infinitely many independent elements? Or do I have to show that $\{b_1,b_2,...\}$ form orthogonal basis?
And I have no idea for part $2)$
Hint: If the only perpendicular vector of a subspace $W$ of an inner product space $X$ is {$0$} then $W=X$ or $W$ is dense in $X$.