Show that there is a unique extension $\overline{T}\in(l^2)'$ Consider the subspace $Y=\{\{x_k\}_k\in l^2:x_k=0\ $ whenever $k$ is odd $\}\subset l^2$ and a linear functional $T\in Y', T\ne 0.$ Show to there is a unique linear extension $\overline{T}\in(l^2)'$ of $T$ with $||\overline{T}||_{(l^2)'}=||T||_{Y'}.$
I know what to do if I have a defined functional $T:W\to \mathbb{R}, \{x_k\}_k\mapsto 2017x_k$, with $W=\{\{x_k\}_k\in l^1, x_k=0, \forall k\in\mathbb{N}, k\ne 2017\}$. In this case I would calculate the dual norm of $T$ and then I would find $\overline{T}$ extension of $T$ with $\overline{T}(\{x_k\})=\sum_1^\infty x_ky_k$ with $\{y_k\}\in l^{\infty}$.
How I can solve this problem without knowing the definition of the functional $T$?
 A: The existence of a extension is guaranteed by the Hahn-Banach-Theorem, but in general this extension is not unique (depending on the Banach space). Here, you have a Hilbert space and in this case you don't need the Hahn-Banach-Theorem and we also get uniqueness of the extension.
Note that $Y$ is closed subspace and thus also a Hilbert-Space. The Riesz representation theorem implies that $T(x) = \langle x,y \rangle$ for a unique $y \in Y$ and $\|T\|=\|y\|$. This gives the unique extension $\overline{T}(x) = \langle x, y \rangle$.
Let us prove that the extension is unique: Any other extension $L \colon H \rightarrow \mathbb{R}$ has a representation $L(x) = \langle x, l \rangle$ with $l \in H$ and $\|l\|=\|y\|$. Decompose $$l = y' + u$$ with $y' \in Y$ and $u \in Y^\perp$. Note that we used the geometry of Hilbert spaces in form of the existence of an orthogonal projection. Then we have for $x \in Y$ that $$\langle x, y \rangle = \overline{T}(x) = T(x) = L(x) = \langle x, y' \rangle.$$
Thus $\langle x, y-y' \rangle =0$. Taking $x=y-y'$ we get that $y=y'$. Now we have
$$\|y\|^2 = \|l\|^2 = \|y'\|^2 + \|u\|^2 = \|y\|^2 +\|u\|^2$$
and thus $u=0$. This proves $l=y$.
