# Boyd & Vandenberghe, problem 3.49 (c) — proving that a function is log-concave

In the problem 3.49(c) of the book it is asked to prove that product over sum function is log-concave. I have some questions related to the solution of this problem which I have mentioned in the following. Please clarify them.

In the solution manual, they have shown that the double derivative of $$\tilde{f}(t)=\sum_{i}\log(x_i+tv_i)-\log \sum_{i}(x_i+tv_i)$$ is less than equal to zero at $$t=0$$. The double derivative of $$\tilde{f}(t)$$ at $$t=0$$ is given as $$\tilde{f}(0)''=-\sum_{i}\frac{v_i^2}{x_i^2}+\frac{(\sum_iv_i)^2}{(\sum_i x_i)^2}$$ which should be $$\leq 0$$ for all $$v$$. It is written in the solution manual that when $$\sum_iv_i\neq 0$$ then $$\tilde{f}(t)=\sum_{i}\log(x_i+tv_i)-\log \sum_{i}(x_i+tv_i)\leq 0 \tag1$$ is homogeneous of degree two in $$v$$ (what does this mean and why it is homogenous of degree two in $$v$$?) and due to this reason they say that it can be assumed that $$\sum_iv_i=\sum_ix_i$$ (why it can be assumed without generality?). After this they say that Eq. (1) is equivalent to showing that $$\sum_{i}\frac{v_i^2}{x_i^2}\geq 1$$holds whenever $$\sum_iv_i=\sum_ix_i$$. To establish this they fix a value of $$x$$ and minimize the convex quadratic form $$\sum_{i}\frac{v_i^2}{x_i^2}$$ over $$\sum_iv_i=\sum_ix_i$$. After this I do not understand how they reach to the conclusion that the optimality condition gives $$\frac{v_i}{x_i^2}=\lambda$$

Please explain this part too. Is there some use of KKT conditions? (Actually, up till chapter 3 the book does not discuss about KKT conditions so I do not understand even if KKT conditions are used.) Any help in this regard will be much appreciated. Thanks in advance.

• Please phrase your question properly, what you claim to be $\tilde{f}(0)$ does not look like $\tilde{f}(0)$. – LinAlg Oct 22 '18 at 1:34

Homogeneous of degree 2 means that if you scale $$v$$ by a factor $$c$$, then the expression $$\tilde{f}''(0)$$ increases by a factor $$c^2$$. Since it is $$c^2$$, the sign of the inequality $$\tilde{f}''(0)\leq 0$$ does not change after rescaling $$v$$, even if $$c<0$$. Therefore, we can always rescale $$v$$ to have its sum equal to the sum of $$x$$.
Then they study the problem: $$\min_v \left\{ \sum_i \frac{v_i^2}{x_i^2} : \sum_i x_i = \sum_i v_i \right\}$$ The Lagrangian is $$L(v,\lambda) = \sum_i v_i^2 / x_i^2 + \lambda(\sum_i x_i - \sum_i v_i)$$, so the stationarity condition (taking the derivative with respect to $$v_i$$) is $$2 v_i / x_i^2 - \lambda = 0$$ for $$i=1,2,\ldots,n$$, leading to $$\lambda = 2v_i/x_i^2$$. The factor 2 is not that relevant, it will cancel out in solving $$\sum_i x_i = \sum_i v_i$$.
Here is an alternative proof since I noticed you asked about $$v$$ and $$t$$ before. The hessian of $$f$$ is: $$H=-\begin{pmatrix} \frac{1}{x_1^2} & c & \cdots & c \\ c & \frac{1}{x_2^2} & \cdots & c \\ \vdots & \vdots & \ddots & \vdots \\ c & c & \cdots & \frac{1}{x_n^2} \end{pmatrix}$$ with all off-diagonal elements $$c=1/(\sum_i x_i)$$. We need it to be negative semidefinite for concavity, so: $$v^THv= - \sum_i \frac{v_i^2}{x_i^2} + \frac{1}{(\sum_i x_i)^2}\sum_{i,j : i \neq j}v_i v_j \leq 0 \quad \forall v.$$ This is also homogeneous of degree 2, so (assuming $$v \neq 0$$) we can rescale $$v$$ to get $$\sum_{i,j : i \neq j}v_i v_j = (\sum_i x_i)^2$$. The remainder of the proof is similar.