# Why can I choose elements of $X$ (and $h \in L^2(0,T)$) in this way? (Dual spaces, norms, Bochner spaces)

This is from the book Vector Measures by Diestel and Uhl, page 98:

Let $X$ be a Banach space. Let $\epsilon > 0$ and suppose first that $g = \sum_{i=1}^\infty x_i^* \chi_{E_i}$ where $x_i^* \in X^*$ and the $E_i$ are a countable partition of $[0,T].$ Choose $h \in L^2(0,T)$ such that $$0 < \lVert h \rVert_{L^2(0,T)} \leq 1$$ and such that $$\lVert g \rVert_{L^2(0,T;X)} - \epsilon < \int_0^T \lVert g(t) \rVert_{X}h(t)$$ and choose $x_i \in X$ with $\lVert x_i \rVert_{X} = 1$ such that $$\lVert x_i^* \rVert_{X^*} -\epsilon \lVert h\rVert_{L^1(0,T)} < x_i^*(x_i)$$

How do I see that I can indeed choose $h \in L^2(0,T)$ and elements $x_i \in X$ in such a fashion? How do I know they exist?

• They write "Suppose first that $g = \sum_{i=1}^\infty x_{i}^\ast \chi_{E_i}$ ...", so the $x_i^\ast$ are given by hypothesis and you only need to choose the $x_i$. For this, use the definition of the operator norm $\lVert x_i^\ast\rVert = \sup_{\lVert x \rVert = 1} |x_i^\ast (x)|$. – Martin May 19 '13 at 9:11
• @Martin If I choose $h$ such that its $L^1$ norm is big enough I can get the final inequality, I suppose. But this may mean that the first inequality does not hold.. – michael_faber May 19 '13 at 9:35
• But $\lVert h\rVert_p \leq 1$, so $\lVert h\rVert_1 \leq T^{1-\frac1p} \lVert h\rVert_p$. – Martin May 19 '13 at 9:49
• Sorry, how is $x_i^*\chi_{E_i}$ meant? $\chi(E_i)$ inputs reals from $[0,T]$ and $x_i*$ inputs elements of $X$.. – Berci May 19 '13 at 10:11
• @Berci Yes, so $g \in X^*.$ Think of what a simple function is. – michael_faber May 19 '13 at 10:12

Take $h(t):=\Vert g\Vert_{L_2(0,T,X)}^{-1}\Vert g(t)\Vert_X$, then you'll get even more strong property: $$\Vert g\Vert_{L_2(0,T,X)}=\int\limits_0^T\Vert g(t)\Vert_X h(t)dt$$ and $\Vert h\Vert_{L_2(0,T)}=1$
To choose $x_i$ note that from definition of the norm of the functional $x_i^*$ we know that for any $\delta>0$ we can find $x_i$ (which depend on $\delta$) with $\Vert x_i\Vert=1$ such that $$\Vert x_i^*\Vert-\delta<x_i^*(x_i)$$ It is remains to take $\delta=\varepsilon\Vert h\Vert_{L_1(O,T)}$
• Thank you. If I wanted $h \in L^p(0,T)$ and $g \in L^q(0,T;X)$ (p,q conjugate), do you know what $h$ should be? Setting $h=\frac{\lVert g(t) \rVert_{X}^{\frac q p}}{\lVert g \rVert_{L^q(0,T;X)}^{{\frac q p}}}$ doesn't give me my second displayed equation (with the $L^q(0,T;X)$ on the $g$ of course) – michael_faber Aug 26 '13 at 11:04
• Never mind, I think it does work with the $h$ as my comment. – michael_faber Aug 26 '13 at 11:29
• For picking the $x_i$, my $x_i$ depends on $\delta$ (as you said) and also on the element $s$ that gives the supremum ($s=\arg \sup_{|x|=1} x_i^*(x)$). That's allowed isn't it? – michael_faber Aug 26 '13 at 15:31
• As for picking $x_i$ recall that $\Vert x_i^*\Vert=\sup\{|x_i^*(x)|:\Vert x\Vert=1\}$. As for the second question, in general $\operatorname{argsup} x_i^*(x)$ may not exist. But you can always approximate it. – Norbert Aug 26 '13 at 15:33
• From the definition of supremum we get that for any $\delta>0$ there exist $x_i'$ with $\Vert x_i'\Vert=1$ such that $\Vert x_i^*\Vert-\delta<|x_i^*(x_i')|$. Now set $x_i=x_i'\operatorname{sign}{x_i^*(x_i')}$ to see that $\Vert x_i^*\Vert-\delta<|x_i^*(x_i')|=x_i^*(x_i)$ – Norbert Aug 26 '13 at 16:34