# uniqueness of Hahn-Banach extension for convex dual spaces

Let $X'$ be strict convex, i.e. for all $x_1',x_2'\in X'$ with $\|x_1'\|_{X'}=\|x_2'\|_{X'}=1$ the implication

$$\left\|\frac{x_1'+x_2'}{2}\right\|=1\Rightarrow x_1'=x_2'$$ holds.

In this case the Hahn-Banach-extension is unique.

I am trying to figure out how I can show this. The Hahn-Banach theorem says that for a subspace $U\subset X$ of a normed space $X$, there exists an extension $x'\in X'$ with $x'|_U=u'$ for every map $u':U\to \mathbb C$.

I've already gone through the proof of the Hahn-Banach theorem, but I don't see where I have to use the convexity of $X'$ to show that the extension is unique.

Can anyone help me here? Thanks.

• See this. Jun 9, 2014 at 11:28

I would like to use slightly different notations. Suppose $h \in U'$ have distinct Hahn-Banach extensions $f$ and $g$. Then $f_1 = \dfrac{f}{||h||}, g_1 = \dfrac{g}{||h||} \in S_{X'}, f_1 \neq g_1$, where $S_{X'}$ is the unit sphere in $X'$. By strict convexity of $X'$, $$\bigg|\bigg|\frac{f_1 + g_1}{2}\bigg|\bigg| < 1,$$ and it follows that $||f + g|| < 2||h||$. However, we have \begin{align*} ||f + g|| &= \sup_{x \in X, ||x|| = 1} | f(x) + g(x) | \\ &\geq \sup_{x \in U, ||x|| = 1} | f(x) + g(x) | \\ &= 2\sup_{x \in U, ||x|| = 1} |h(x)|\\ &= 2||h||, \end{align*} a contradiction.

• Thanks, but I don't quite understand why $f_1,g_1\in S_{X'}$. Since $f,g$ are extensions of $h$, we have $f_1,g_1\in S_{U'}$, but why do we also have $f_1,g_1\in S_{X'}$? Jun 9, 2014 at 13:57
• Please note that Hahn-Banach extension has two properties: 1) $f$ and $g$ coincide with $h$ on $U$, namely $f|_U = g|_U = h$. 2) Norm preserving, which means $||f||_{X'} = ||g||_{X'} = ||h||_{U'}$. So $||f_1||_{X'} = \bigg|\bigg| \frac{f}{||h||_{U'}} \bigg|\bigg|_{X'} = \frac{||f||_{X'}}{||h||_{U'}} = 1$. Jun 9, 2014 at 14:03
• oh, right, I totally forgot about that. Thanks! Jun 9, 2014 at 14:21