Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello, everyone!

As we know that by Jensen's inequality, for jointly convex function $f$ and $\sum_ix_i^2=1$, we have $$f(\sum_i{x_i^2\lambda_i},\sum_i{x_i^2\theta_i)}\leq\sum_i{x_i^2f(\lambda_i,\theta_i)}\leq\max_if(\lambda_i,\theta_i)\leq\sum_if(\lambda_i,\theta_i),$$ and we get a bound of $f(\sum_i{x_i^2\lambda_i},\sum_i{x_i^2\theta_i)}$ independent of $\{x_i\}$.

However, I wonder if this inequality can be extended to the case where the probability distribution $\{x_i\}$ on the two variables of $f$ are not identical but just constrained.

To be more specifically, suppose that $f:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ is jointly convex with both its arguments, $V=[\mathbf{v}_1\;\mathbf{v}_2\;\ldots\;\mathbf{v}_m]$ and $U=[\mathbf{u}_1\;\mathbf{u}_2\;\ldots\;\mathbf{u}_m]$ are orthogonal matrices and thus $\{\mathbf{v}_i\},\{\mathbf{u}_j\}$ consist an orthonormal basis in $\mathbb{R}^m$ respectively.

Then for any $\mathbf{x}\in\mathbb{R}^m$ satisfying $\|\mathbf{x}\|=1$, I wonder if there is a relationship between $L_1$ and $L_2$ shown in the following two formulas. \begin{eqnarray} L_1&=&f\left({\sum_i(\mathbf{v}_i^\top\mathbf{x})^2\lambda_i},\sum_j{(\mathbf{u}_j^\top\mathbf{x})^2\theta_j}\right)\\\ L_2&=&\sum_i\sum_j{(\mathbf{v}_i^\top\mathbf{u}_j)^2f(\lambda_i,\theta_j)} \end{eqnarray}

In language of matrix, $L_1$ can also be formulated as $f\left(\mathbf{x}^\top V\Lambda V^\top\mathbf{x},\mathbf{x}^\top U\Theta U^\top\mathbf{x}\right)$.

Considering that $\sum_i(\mathbf{v}_i^\top\mathbf{x})^2=\sum_j(\mathbf{v}^\top\mathbf{x})^2=1$, my question is that does there exist an inequality about $L_1$ and $\gamma L_2$ where $\gamma$ is any constant independent of $\mathbf{x}$?

How should I consider about this problem? Or are there any papers about this topic for reference?

Could anyone be so kind to help me about this question? Any suggestion will be appreciated! Thank you very much!


Remark:

I tried to simply apply the Jensen's inequality to $L_1$ and get the result $$L_1\leq\sum_i\sum_j{(\mathbf{v}_i^\top\mathbf{x})^2(\mathbf{u}_j^\top\mathbf{x})^2f(\lambda_i,\theta_j)}.$$ Does there exists any relationship between this formula and $L_2$?

Any suggestion will be appreciated! Thank you very much!

share|improve this question
2  
Please do not use too many math symbols in the title of a question. It slows down the main site. –  Marc Palm Apr 2 '12 at 18:03
    
But: been there, done that as well;) –  Marc Palm Apr 2 '12 at 18:05
    
@Marc, OK! I will remember this next time. Thanks! –  ppyang Apr 3 '12 at 2:29

1 Answer 1

up vote 2 down vote accepted

Consider the following case. Let $f(a,b) = a + b$, $\lambda_1 = \theta_1 = -1$, $\lambda_2 = \theta_2 = 1$, and $V = U = I_2$. Then, $-2 \leq L_1(\mathbf x) \leq 2$ and $L_2 = 0$. Clearly, there exists no $\gamma\in\mathbb R$ that satisfies either $L_1(\mathbf x) \leq \gamma L_2$ or $L_1(\mathbf x) \geq \gamma L_2$ for all $\mathbf x\in\mathbb R^2$ such that $\|\mathbf x\| = 1$.

share|improve this answer
    
Thank you for your counter example! –  ppyang Apr 3 '12 at 10:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.