Let $X$ be the vector space $\mathbb{R}^2$ and let $Y$ be the subspace of $X$ consisting of all $(\alpha, \beta)$ such that $\beta = 0.$ Define $ f_0 = Y \rightarrow \mathbb{R} $ by the formula $f_0(\alpha,0)=\alpha. $ Notice that $f_0$ is a bounded linear functional on $Y$ with respect to any norm given to $X$.

(a) Suppose that $X$ is given the Euclidean norm. Show that $f_0$ has a unique Hahn-Banach extension to $X$.

(b) Suppose that $X$ is given the norm that makes it into $l_1^2$. Show that $f_0$ has an infinite number of different Hahn-Banach extensions to $X$.

Attempted proof

(a) If $f_0$ is to have any linear extension $f: \mathbb{R}^2 \rightarrow \mathbb{R} $, this extension should look like $f(x,y)=ax + by $ for all $(x,y) \in \mathbb{R}^2$. Thus $\lVert f \rVert = \sqrt{a^2 + b^2} $ (though I'm not quite sure about this. A quick argument for it will be appreciated). By the Hahn-Banch Theorem, since the operator norm of $f_0$ should equal the operator norm of $f$, we must have \begin{equation} 1 = \sqrt{a^2 + b^2} \end{equation} Again, by the Hahn-Banch Theorem, since the restriction of $f$ to $Y$ should equal $f_0$ we must have for arbitray $x$: $$f\lvert_Y = f_0 \Rightarrow ax+by\rvert_{y=0} = x \Rightarrow ax = x\Rightarrow a=1 $$ It follows that $b=0$, and since this solution is unique, we've found the uniqe extention to $X$, which is $f(x,y) = x$ for all $(x,y) \in \mathbb{R}^2$.

For part (b) I have absolutely no idea. Any help will be appreciated.

  • 1
    $\begingroup$ Your proof is OK. For (b), note that $f$ has norm $1$, so any extension $\hat{f}$ must have the form $\hat{f}(x,y)=x+by$. For which $b$ $\hat{f}$ has norm $1$? Note that this time the norm of $\hat{f}$ is computed respect to the $l_1^2$-norm $\endgroup$ – Veridian Dynamics Jan 13 '18 at 23:14
  • $\begingroup$ @Calvin What is it then? $\endgroup$ – Erfan Jan 13 '18 at 23:51
  • $\begingroup$ My apologies, it is actually correct. $\endgroup$ – Calvin Khor Jan 13 '18 at 23:56
  • $\begingroup$ @Calvin No worries :)) $\endgroup$ – Erfan Jan 14 '18 at 6:36
  • $\begingroup$ @Veridian Thanks for the insightful comment $\endgroup$ – Erfan Jan 14 '18 at 7:49

For $(a)$, \begin{align} \|f\|&=\sup\{|ax+by|:\ x^2+y^2=1\}. \end{align} By Cauchy-Schwarz, $$|ax+by|\leq\sqrt{a^2+b^2}\,\sqrt{x^2+y^2},$$ and the equality is achieved. So $\|f\|=\sqrt{a^2+b^2}$.

For part $(b)$, we still have $f(x,y)=ax+by$. But now $$ \|f\|=\sup\{|ax+by|:\ |x|+|y|=1\}=\max\{|a|,|b|\}. $$ So $f_b(x,y)=x+by$ is a Hahn-Banach extension for all $b$ with $|b|\leq1$.

  • $\begingroup$ Thanks :) Its all clear now $\endgroup$ – Erfan Jan 14 '18 at 7:48

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