Let $G=\{(x_n)\in l_1 : x_1 - 3x_2 = 0\}$ and $f : G\to \mathbb{R}$ be defined by $f(x) = x_1$. Prove that $g(x) = (3/4)(x_1+x_2)$ is the unique Hahn-Banach extension for $f$.

(I found $g$ is an extension for $f$, $||g||=3/4$. But how to prove $g$ is unique. By Hahn-Banach theorem, for a bounded linear functional $f$ defined on a subspace $G$ there exists an extension $g$ with $||f||=||g||$.)


Observe, that $||f||=\frac34$. Indeed, $$G=\{(x_1,\tfrac13x_1,x_3,x_4,\dots)\in\ell_1\},$$ and $\max_{x\in G,||x||=1}|f(x)|$ is achieved at the vector $x=(\tfrac34,\tfrac14,0,0,0,\dots)\in G$.

Hence, we are looking for an extension $g$ such that $||g||\leq\frac34$ (in fact $||g||=\frac34$).

We extend $f$ to $g$ by setting $g(a,0,0,\dots)=\lambda a,\ a\in\mathbb R$. Then $|\lambda|$ must be $\leq\frac34$. Further, $$g(0,a,0\dots)=g(3a,a,0\dots)-g(3a,0,0\dots)=f(3a,a,0\dots)-g(3a,0,0\dots)=3a-3a\lambda,$$ for all $a\in \mathbb R$. Hence $|3-3\lambda|$ must be also $\leq \frac34$. Combining the inqualities $|\lambda|\leq \frac34$ and $|3-3\lambda|\leq\frac34$ we obtain $\lambda=\frac34,$ i.e. the Hahn-Banach extension is unique.

Remark. The same argument works also over complex numbers.


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