# Is there relationship between $f\left({\sum_i(\mathbf{v}_i^\top\mathbf{x})^2\lambda_i},\sum_j{(\mathbf{u}_j^\top\mathbf{x})^2\theta_j}\right)$ and $\sum_i\sum_j{(\mathbf{v}_i^\top\mathbf{u}_j)^2f(\lambda_i,\theta_j)}$ if $f$ is jointly convex?

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 $\mathbf{x}$?

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!

• 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

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 $\|\mathbf x\| = 1$.