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I am following the the book "Lecture notes on functional analysis". The following extension thm has been proven in the book. enter image description here

If we consider the finite dimensional case of the $X$, does the following prove work for the existence of the linear functional $F$? $\to$ My guess is yes but I have some problem of proving the property $\|F\|=\|f\|$.

Let $\{x_1,\dots, x_k\}$ be basis for $V$. We can extend it to a basis $\{x_1,\dots, x_k,x_{k+1},\dots , x_n\}$ of $X$ and set $F(x_m)=0$ for $m = k+1, \dots , n$. So for any $X\ni x = a_1x_1+\dots+a_nx_n$ , we can define $F(x)=a_1f(x_1)+\dots+a_kf(x_k)$. Then $F$ is a linear functional on $X$, $F(x)=f(x) , \forall x\in V$, and $\|F\|=\|f\|$.

domath
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  • https://math.stackexchange.com/questions/1060848/hahn-banach-proof-by-extension-of-basis – cmk Feb 04 '20 at 21:01
  • I found this http://www.math.unl.edu/~s-bbockel1/928/Hahn_Banach_Theorem.html but the example is not clear to me. – domath Feb 05 '20 at 05:02
  • can the following also be considered to show. I think we do can have such an extension $F$ but the thing is that it would necessarily make $|F|$ comparable to $|f|$.

    $$|F(x)| = |\sum_{i=1}^{k} \alpha_i f(x_i)| \le \sum_{i=1}^{k} |\alpha_i f(x_i) | \le |\alpha| |\sum_{i=1}^{k} | f(x_i)||$$

    $|\alpha|$ still could be large even if $|x|$ is small.

     – domath Feb 05 '20 at 05:04
  • Yes, as well as the link I added after the comment. – domath Feb 10 '20 at 02:26

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