3
$\begingroup$

Let $X$ be a real normed linear space, and $l$, $l_1$ linear functionals on a subspace $Y$ of $X$, such that $$|l(y)|+|l_1(y)|\le \|y\|, \quad y\in Y.$$

I would like to show that there exists $L, L_1 \in X^*$ that extend $l$ and $l_1$, respectively, and also $$|L(x)|+|L_1(x)|\le \|x\|, \quad x\in X.$$

I know that the extension can be made by the Han-Banach theorem, since $l, l_1$ are bounded functionals. However, I am not sure how I can guarantee that the dominance condition on the whole vector space still holds. Any thought?

$\endgroup$
  • 1
    $\begingroup$ Thought I had it. Good question $\endgroup$ – Patrick Da Silva Mar 15 '13 at 20:17
  • $\begingroup$ The question is equivalent to asking if Hahn-Banach works when the image is $\mathbb R^2$ equipped with the $1$-norm. I don't think this is trivial. $\endgroup$ – Patrick Da Silva Mar 15 '13 at 20:26
  • $\begingroup$ @PatrickDaSilva Like with Hahn-Banach, this is easy if the domain is in a Hilbert space. But for a general normed vector space... I had never thought of that. $\endgroup$ – Julien Mar 15 '13 at 20:36
  • $\begingroup$ I found something: the Hahn-Banach-Kantorovich theorem. See here. I have not yet figured out if this helps. $\endgroup$ – Julien Mar 15 '13 at 20:45
  • 1
    $\begingroup$ @julien : I'm not sure it does... it says "ordered linear spaces", which doesn't seem to be the case of $\mathbb R^2$. I can't help but notice this is homework, your teacher seems mean... $\endgroup$ – Patrick Da Silva Mar 15 '13 at 20:58
2
$\begingroup$

Define $l_{\pm}=\frac{l\pm l_1}{2}$ on $Y$. Then $$ |l_\pm y|\le \frac{1}{2}(|ly|+|l_1y|)\le \frac{1}{2}\|y\| \quad y\in Y. $$ Extend $l_\pm$ by the Hahn-Banach theorem to $L_\pm$ on $X$, then $$ |L_\pm x| \le \frac{1}{2} \| x\|, \quad x\in X. $$ Let $ Lx=L_+x+L_-x$ and $L_1x=L_+x-L_-x$ for $x\in X$, and observe that \begin{align} L|_Y&=l_++l_-=\frac{l+l_1}{2}+\frac{l-l_1}{2}=l\\ L_1|_Y&=l_+-l_-=\frac{l+l_1}{2}-\frac{l-l_1}{2}=l_1 \end{align} and $$ |Lx|+|L_1x|=|L_+x+L_-x|+|L_+x-L_-x| \le \max\{2|L_+x|, 2|L_-x|\}\le\|x\|. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.