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.
Best Answer
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.