# Analytical Proof for the Inequality of Minimum Values of a Transcendental Function with Varying Number of Variables

Consider the following function:

$$f_k(p_1,p_2,..,p_k) = \frac{2}{1 + d}\left((-1)^k + 2\sum_{i=1}^k (-1)^{i+1}p_i\right)\left((-1)^k + 2\sum_{i=1}^k (-1)^{i+1}p_i^{d+1}\right)$$

where $$k$$ is the number of variables, and the following conditions hold:

• $$d > 0$$
• $$0 < p_i < 1$$ for all $$i = 1, 2, \ldots, k$$
• $$p_i < p_{i+1}$$ for all $$i = 1, 2, \ldots, k-1$$
• $$k \in \mathbb{N}$$

I suspect that the following inequality is true:

$$\min_{p_1, p_2, \ldots, p_k} f_{k \geq 2} > \min_{p_1} f_{k=1}$$

Meaning that the minimum value of the function $$f$$ with more than 2 variables is always greater than the minimum value of function with one variable. As an example, consider the following:

$$f_1(p_1) = \frac{2}{1 + d}\left(-1 + 2p_1\right)\left(-1 + 2p_1^{d+1}\right)$$

$$f_2(p_1,p_2) = \frac{2}{1 + d}\left(1 + 2(p_1 - p_2)\right)\left(1 + 2(p_1^{d+1} - p_2^{d+1})\right)$$

Now I suspect that the following inequality is true:

$$\min_{p_1, p_2} f_2(p_1,p_2) > \min_{p_1} f_1(p_1)$$

I have confirmed this numerically, but I am struggling to validate it analytically. The problem arises from the fact that the derivative of this function (with respect to $$p_i$$) even for $$k=1$$ is a transcendental equation with no analytical solution. Therefore, the obvious approach of setting the derivative to zero is not possible. I am wondering if there is a way to exploit the given conditions to prove the stated inequality. Any help would be greatly appreciated.

• $f$ seems to depend on $k$ and $p_1,\dots,p_k$ I find your notation very confusing. It is not at all oobvious what the minimums are over. You can't even have $p_1,\dots,p_k$ until you know $k,$ since $k$ is the number of elements. Commented Apr 27 at 0:29
• I've changed parts of my notation and rephrased my intent, is it better now? @ThomasAndrews Commented Apr 27 at 0:37
• I also added an example. Please let me know if it is still unclear. @ThomasAndrews Commented Apr 27 at 0:48
• It's not clear why you call these functions "transcendental." They look like polynomials. Commented Apr 27 at 1:23
• If you index the $p_i$ in descending order instead of ascending, I believe you can write $f$ in the more readable form $f_k(p_1, ..., p_k) = 2/(1+d)\left[1 +2 \sum_{i=1}^k (-1)^i p_i\right]\left[1 + 2\sum_{i=1}^k (-1)^i p^{1+d}_i\right]$ Commented Apr 28 at 17:12

The method I used in my answer to the $$k = 2$$ case can be generalized to the $$k \ge 2$$ case to show that at any critical point, the Hessian has a positive and a negative eigenvalue, and thus cannot be a local extremum. Indexing the $$p_i$$ so that $$p_i > p_{i+1}$$ allows for a simpler representation of the function, $$f_k(\mathbf{p}) = \frac{2}{1+d}\left[1 + 2\sum_{n = 1}^k(-1)^np_n\right]\left[1 + 2\sum_{n = 1}^k(-1)^np_n^{d+1}\right].$$ The partial with respect to $$p_i$$ is $$\partial_i f_k = (-1)^i4\left(\frac{1}{1+d}\left[1 + 2\sum_{n = 1}^k(-1)^np_n^{d+1}\right] + p_i^d\left[1 + 2\sum_{n = 1}^k(-1)^np_n\right]\right).$$ At a critical point, every one of these partials must be $$0$$. So, in particular, $$\partial_i f + \partial_{i+1}f$$ will also be zero. This allows us to cancel the first term in the expression and get $$\partial_i f_k + \partial_{i+1} f_k = 4(-1)^i\left(p_i^d-p_{i+1}^d\right)\left[1 + 2\sum_{n = 1}^k(-1)^np_n\right] = 0$$ Since by assumption $$p_i > p_{i+1}$$ in the domain, the only way for this to vanish is that $$1 + 2\sum_{n=1}^k(-1)^n p_n = 0$$. Thus, any critical point in the domain must satisfy that condition.

Taking the second derivative then gives $$\partial_i\partial_j f_k = (-1)^{i+j}8\left(p_j^{d}+ p_i^d + \delta_{ij}\frac{d}{2}p_i^{d-1}\left[1 + \sum_{j = 1}^k(-1)^jp_j\right]\right)$$ The last term vanishes at any critical point in the domain, so the Hessian at a critical point will be $$H_{ij} = (-1)^{i+j}8(p_i^d + p_j^d).$$ This is a rather simple rank 2 matrix whose eigenvalues can be readily found. There will only be two nonzero eigenvalues, $$\lambda_{\pm} = \sum_{n=1}^k p_n^d \pm 2\sqrt{k\sum_{n=1}^k p_n^{2d}}.$$ Obviously $$\lambda_+$$ is positive, while $$\lambda_-$$ is seen to be negative from the fact that the RMS is always greater than the mean. Thus, no critical point of $$f_k$$ in the domain can be a local extremum and its global minimum must lie on the boundary.

As this answers part of the question and there are lots of comments, I thought it would be better to make it a post. @eyeballfrog was able to find a solution for the $$k=2$$ case, showing that it doesn't have a global minimum:

Proving that a function doesn't have global minimum in a specific domain

Meaning that we have to check the boundaries, and by doing so, minimum happens when:

$$p_2 \rightarrow 1$$

which converts the $$f_2$$ back to $$f_1$$. However, the method mentioned by him doesn't apply for $$f$$ with more than two variables, so the problem still remains unsolved.