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Given a Banach space $X$ and its dual space $X^*$. Suppose $a$ and $b$ are two unit norm vectors in $X$. $a^*$ and $b^*$ be unit norm elements in dual space $X^*$ such that $a$ and $a^*$ as well as $b$ and $b^*$ are aligned. Is $\langle a,b^*\rangle = \langle b,a^*\rangle$? If yes please present a proof else a counter example.

Note:$\langle x,x^*\rangle$ is used to denote the action of dual space elements($x^*\in X^*$) on the elements in the original space($x\in X$).

$a$ aligned with $a^*$ $\Rightarrow \langle a,a^*\rangle = \|a\|\|a^*\| = 1$. Similarly $\langle b,b^*\rangle = \|b\|\|b^*\| = 1$.

According to me in a Hilbert space setting $\langle a,b^*\rangle = \langle b,a^*\rangle$ is equivalent to $\langle a,b\rangle = \langle b,a\rangle$ or equivalently $a^Tb = b^Ta$. Thus it holds for Hilbert spaces owing to the symmetric nature of inner product in a Hilbert space. So I believe that the result might hold in a Banach space.

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  • $\begingroup$ You write that $a^{*}$ is the element of $X^{*}$ of unit length aligned with $a$. I don't think there's necessarily only one. For example, consider $a = (1,1)$ in $\mathbf{R}^2$ with the sup norm. Any line through $(1,1)$ with negative slope corresponds to an $a^{*}$ that is aligned with $a$. $\endgroup$
    – user180040
    Oct 1, 2014 at 2:00
  • $\begingroup$ You are correct about the fact that there could be multiple $a^*$ possible. Even if you consider any one of them does the condition $<a,b^*> = <b,a^*>$ hold. $\endgroup$
    – saurav90
    Oct 1, 2014 at 2:06
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    $\begingroup$ I suggest using \langle and \rangle in place of < and > for inner products. It is much easier on the eyes. $\endgroup$ Oct 1, 2014 at 2:08
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    $\begingroup$ Thanks @Cameron Williams . Suggestion has been implemented. $\endgroup$
    – saurav90
    Oct 1, 2014 at 3:13

1 Answer 1

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Consider $\mathbf{R}^2$ with the sup norm. Define $a=(1,0)$, $a^{*}(x,y) = x$, $b=(1,1)$, $b^{*}(x,y) = y$. Then $a$ and $a^{*}$ are aligned, as well as $b$ and $b^{*}$. However $\langle a,b^{*} \rangle = 0$ and $\langle b,a^{*} \rangle = 1$

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  • $\begingroup$ Could you also think of a counter example using other norms apart from the sup norm? Also any comments on conditions under which the result would hold is of great help. $\endgroup$
    – saurav90
    Oct 1, 2014 at 15:18
  • $\begingroup$ I don't know. You could ask the same question again in the case of a strictly convex space. Also, what are sufficient conditions for $a^{*}$ to be unique? $\endgroup$
    – user180040
    Oct 2, 2014 at 0:00

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