I’ve encountered the following assumption:

Let D be a set such that there exists a Minkowski function $f(u)$ on $\mathbb{R}^l$ and norm $g(v)$ on $\mathbb{R}^m$ such that $\forall u\in \mathbb{R}^l, \forall v\in \mathbb{R}^m$ $ \max_{X\in D} u^{t}Xv = f(u)g(v)$.

According to the author, there are many functions which accomplish this assumption.

One example is the set $D=\left( X| \Vert{X\Vert}_F \leq 1 \right)$ for which $f(u)=\Vert u \Vert_2$, $g(v)=\Vert v\Vert_2$. The upper bound $\max_{X\in D} u^{t}Xv \leq f(u)g(v)$ is easy to show by applying the Cauchy Schwartz inequality and then the consistency of the Frobenius norm. However, I can’t find an example to show a member in the set D attains an equality. That is, that $ \exists X\in D ,\ u^{t}Xv = f(u)g(v)$.

Can anyone help prove this assumption for this set (or any other)?


  • $\begingroup$ @Mark: I deleted my answer because it looked ugly once one writes out the final answer; I wrote it using that notation to show that everything was essentially a vector. $\endgroup$
    – Suvrit
    Oct 28 '10 at 20:45

Thanks to Marks replay (which he deleted) I realized the answer to my question (without vectorization of the problem, so notation stays simple). Choose $X^{*}=uv^{T}/\Vert u \Vert_2 \Vert v \Vert_2$ :

$\Vert X^{*} \Vert_F =1/\Vert u \Vert_2 \Vert v \Vert_2 * \Vert uv^T \Vert_F = 1/\Vert u \Vert_2 \Vert v \Vert_2 * \Vert u \Vert_2 \Vert v \Vert_2=1 $ so $X^{*} \in D$. (The last equality can be easily verified by opening the Frobenius norm).

The inner product is then:

$u^TX^{*}v = u^T uv^T v/\Vert u \Vert_2 \Vert v \Vert_2 = \Vert u \Vert_2 \Vert v \Vert_2 $.

  • $\begingroup$ I didn't delete my reply. I wrote a comment to Suvrit's answer, which Suvrit subsequently deleted. $\endgroup$ Dec 12 '10 at 1:56

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