A question regarding Hahn Banach theorem

Question

Let $Y=\{(a_n)\in l^2 : a_2=2a_1 , a_4=4a_3\}$. Is there exist a functional $f \neq 0 \in X^*$ such that $f(Y)={0}$.

I know that it can be done using the Hahn-Banach corollary, which says, if $X$ is a normed space, $Y\subset X$ a closed sub-space, and $x \in X\setminus Y$, then there is a functional $f \in X^*$ such that $\|f\|=1$ , $f$ restricted to $Y$ equals $0$ ,and $f(x)=d=\inf_{y\in Y} \|x-y\|$.

Clearly $l_2$ is a normed space. And we can find a sequence $a_n\in l^2\setminus Y$ with $\|a_n\|_2=1$. In order to show that $Y$ is closed in $X$ I have to take a sequence $b_n\in Y$ such that $\lim_{n\to \infty} b_n =x$ and show that $x\in X$ but is not that obvious according to $Y$'s definition?

Any support would be kindly welcomed.

Answer 1

Rather than resort to using the Hahn-Banach theorem to solve this problem, perhaps consider a constructive approach:

Define $f:l_2\to\mathbb C$ by $f(a_n)=-2a_1+a_2$. This map is certainly linear, $f\neq 0$, $f(Y)=0$, and by Hölder's inequality, we have $$|f(a_n)|=|-2a_1+a_2|\leq((-2)^2+1^2)^{1/2}\|(a_n)\|_2=\sqrt 5\ \|(a_n)\|_2,$$ So $f\in(l_2)^*$.

Answer 2

Since norm-convergence in $\ell^2$ implies pointwise convergence (each entry of the sequence converges), it follows that $Y$ is a closed subspace of $\ell^2$. Moreover, it is easy to see that $\ell^2 \neq Y.$ Then invoke the following corollary of the Hahn-Banach (separation) theorem:

Let $V$ be a normed space and $Y$ be a proper closed subspace of $V$. Then there exists a non-zero continuous functional $\tau \in V^*$ with $\tau(Y) = \{0\}$.