In one of the problem (3.49(c)) of Convex Optimization book (By Boyd and Vandenberghe) it is asked to show that product over sum function $$g(x)=\frac{\prod_{i=1}^nx_i}{\sum_{i=1}^nx_i}$$ is log concave. Now $f(x)=\log(g(x))$ is given as follows $$f(x)=\sum_{i=1}^n\log(x_i)-\log(\sum_{i=1}^nx_i).$$ Now if $g(x)$ is log concave then $f(x+tv)$ should be concave for all values of $t$ which means that the double derivative of $f(x+tv)$ with respect to $t$ should be negative for all values of $t$. The double derivative of $f(x+tv)$ with respect to $t$ is given as follows $$f''(x+tv)=-\sum_{i=1}^n\frac{v_i^2}{(x_i+tv_i)^2}+\frac{(\sum_{i=1}^nv_i)^2}{(\sum_{i=1}^nx_i+t\sum_{i=1}^nv_i)^2}.$$ After this the solution manual proves that $f''(x+tv)<0$ for $t=0$. I do not know why they only check for $t=0$ case and ignore all the other values of $t$. Please help me in understanding this part. Thanks in advance.
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Every term in that sum is a square of a number and is therefore positive... – player100 Sep 19 '18 at 01:10
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@player100 please elaborate. The first term is negative and the second term is positive so how can it be always positive or negative? – Frank Moses Sep 19 '18 at 01:18
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Sister question – Rodrigo de Azevedo Apr 09 '21 at 11:35
1 Answers
At any given point $x$, they are taking an aribitrary $1$-dimensional slice of the domain through $x$: the line of points of the form $x + tv$, where $t$ ranges over $\mathbb{R}$. They then show that the second derivative is negative, at that point, along any arbitrary line. They have to let $t$ range freely initially, so that the function is defined along the line, at least in a neighbourhood of $x$, so that differentiation is meaningful.
They don't have to show it for other values of $t$, because they've already shown this for arbitrary $x$. If you want to know whether this holds true for $x + t_0 v$, for some fixed $t_0 \in \mathbb{R} \setminus \lbrace 0 \rbrace$, then simply redefine $x_0 = x + t_0 v$, consider the function $t \mapsto f(x_0 + tv)$, and do all the above steps, including considering where $t = 0$.
What this means is that every $1$-dimensional slice of the function is log-concave. Since concavity is defined along line segments, this means the entire function is log-concave.
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Thank you so much for your answer. It makes much sense to me. But does this means that we can always put $t=0$ for checking concavity/convexity of any other general function? – Frank Moses Sep 19 '18 at 01:20
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Well, as I said at the end of the first paragraph, we need to let $t$ range over $\mathbb{R}$ in order to take a one-dimensional slice of the function. We need an actual function of $t$ to differentiate, which means it has to be defined at more than just $t = 0$. Once we've gotten our derivatives, then we can always discard the other values, and just consider $t = 0$. – Theo Bendit Sep 19 '18 at 01:24
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