# Extending two bounded linear functionals whose sum is dominated by the vector norm

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?

• Thought I had it. Good question – Patrick Da Silva Mar 15 '13 at 20:17
• 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. – Patrick Da Silva Mar 15 '13 at 20:26
• @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. – Julien Mar 15 '13 at 20:36
• I found something: the Hahn-Banach-Kantorovich theorem. See here. I have not yet figured out if this helps. – Julien Mar 15 '13 at 20:45
• @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... – Patrick Da Silva Mar 15 '13 at 20:58

## 1 Answer

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 $$|Lx|+|L_1x|=|L_+x+L_-x|+|L_+x-L_-x| \le \max\{2|L_+x|, 2|L_-x|\}\le\|x\|.$$