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I have a formula containing two squared root components as follows:

$1 - \frac{\sqrt{\sum_{\forall i \in \mathcal{I}}\sum_{\forall j \in \mathcal{J}}(a_{ij} - b_{ij})^2}}{\sqrt{\sum_{\forall i \in \mathcal{I}}\sum_{\forall j \in \mathcal{J}}a_{ij}^2}}$.

Assume that $a_{ij}, \forall i \in \mathcal{I}, \forall j \in \mathcal{J}$ are constants. How to know that the above equation is concave or convex?

bnbfreak
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1 Answers1

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We can view it as $$1-\frac1u \|A-B\|_F$$ where $u$ is a positive constant.

since we know that the frobenius norm is convex, multiplying by a negative number makes it concave. Shifting it by $1$ doesn't change the property.

Hence it is a concave function.

Siong Thye Goh
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  • So is $u$ the same as $\sqrt{\sum_{\forall i \in \mathcal{I}}\sum_{\forall j \in \mathcal{J}}a_{ij}^2}$? I forget that the denominator is also the Frobenius norm but also constant. – bnbfreak Mar 05 '20 at 06:12
  • yes, we are told that it is a constant. – Siong Thye Goh Mar 05 '20 at 06:13
  • Could you help me to answer the edited formula in the above question? Thank you. – bnbfreak Mar 11 '20 at 06:30
  • seems to be true from triangle inequality. – Siong Thye Goh Mar 11 '20 at 06:54
  • Do you mean because $1 - \frac{\bigg|\mathbf{a} - \frac{\sum_{m=1}^M m(x_{m} - x_{m-1})}{\gamma} \mathbf{b}\bigg|F}{|\mathbf{a}|_F} = 1 - \frac{|\mathbf{a}|_F - \frac{\sum{m=1}^M m(x_{m} - x_{m-1})}{\gamma} |\mathbf{b}|_F}{|\mathbf{a}|_F}$? – bnbfreak Mar 11 '20 at 07:02
  • I mean we can prove it from the definition. During the proof, we can use the triangle inequality to prove concavity. – Siong Thye Goh Mar 11 '20 at 07:32