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If $X$ is the Euclidean space $\mathbb{R}^3$ and $f(x)=a_1 x_1 + a_2 x_2$, $x=(x_1, x_2) $ a bounded linear functional on the subspace $\mathbb{R}^2$ of $X$, how do I find a bounded linear functional $f'$ that is an extension of $f$ with the same norm as that of $f$?

Does the functional $f'(x) = a_1 x_1 + a_2 x_2 + 0 \cdot x_3$ work?

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    $\begingroup$ Yes, this works. This space is simple so your function that extends should be simple too, but it's a fair question. $\endgroup$
    – Squirtle
    Commented Nov 24, 2015 at 7:10

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Yes, it works at Squirtle points out in his comment. The Hahn-Banach theorem guarantees you that if $||f||=M$ then $||f'||=M$ too. Considering the Euclidean norm you have:

$$||f(x)||=\sqrt{(a_1x_1)^2+(a_2x_2)^2} \le |a| \sqrt{x_1^2 + x_2^2} = |a| ||x|| $$

Where $a=max\{a_1,a_2\}$ and the equality holds considering $x=(1,0)$ or $x=(0,1)$. If you consider an extension like $f'(x)=a_1x_1+a_2x_2 + a_3 x_3$ the norm must be $|a|$ also, so $a_3 \le |a|$ gives you a condition for the possible valid extensions of $f$.

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