Re-visiting an inequality problem involving convexity

Question

This question is in link with the following question: An Inequality Problem Involving Convexity : that I already asked in math.stackexchange a few weeks back. Lee David Chung Lin gave a very nice counterexample to that question, which was indeed extremely helpful. However, I added a comment there, to which I did not get a reply from anyone yet. In this post, I ask for a specialized (and probably somewhat easier) version of my original question in that comment.

So, here I define a function $f(x) = x \log x + (1-x) \log (1-x)$ for $x \in [0,1]$ (with the obbvious convention that $0 \log 0 = 0$). Suppose that $x_1, x_2 \in [0,1]$, and I define $$\bar{x} = \sqrt{\frac{x_1^2 + x_2^2}{2}}~.$$ Then, I strongly suspect that the following inequality holds: $$f(x_1) + f(x_2) \geq 2 f(\bar{x})~,$$ and further, equality holds if and only if $x_1 = x_2$. Can someone help me prove this inequality, or give a counterexample, if possible? I have checked this inequality using a very fine mesh of $[0,1]^2$ in MATLAB, and so, I believe that this inequality is indeed true.

Any help will be highly appreciated.

Answer 1

This is an elaboration on the answer given by gerw:

Define $$ y_1=x_1^2\\ y_2=x_2^2 $$ and the function $g(y)=f(\sqrt{y})$. Then $$ f(x_1)=g(y_1)\\ f(x_2)=g(y_2)\\ f(\overline{x})=g\left(\frac{y_1+y_2}{2}\right) $$ Then the inequality is simply expressing the convexity of $g$ wrt. $y$: $$ f(x_1)+f(x_2)\geq 2f(\overline{x})\\ \Updownarrow\\ \frac{g(y_1)+g(y_2)}2\geq g\left(\frac{y_1+y_2}{2}\right) $$

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So let us define $f$ a bit more generally as related to your original question: $$ f(x)=x\log\left(\frac xa\right)+(1-x)\log\left(\frac{1-x}b\right) $$ still keeping the definition $g(y)=f(\sqrt y)$. To find the derivatives of $g$ in order to possibly show convexity (depending on $a,b$) we may break this into parts applying the chain rule a couple of times: $$ \left(x\log\left(\frac xa\right)\right)'=\log\left(\frac xa\right)+1 $$ which by applying the chain rule and $(1-x)'=-1$ gives us $$ \left((1-x)\log\left(\frac{1-x}b\right)\right)'=-\log\left(\frac{1-x}b\right)-1 $$ showing that $$ f'(x)=\log\left(\frac xa\right)-\log\left(\frac{1-x}b\right) $$ and since $(\sqrt y)'=1/(2\sqrt y)$ this gives us $$ g'(y)=\frac{\log\left(\frac{\sqrt{y}}a\right)-\log\left(\frac{1-\sqrt{y}}b\right)}{2\sqrt{y}} $$

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Now to find the second derivative, let us then consider the function $$ h(x)=\frac{\log\left(\frac{x}a\right)-\log\left(\frac{1-x}b\right)}{2x} $$ and the derivatives of the subexpressions: $$ \left(\log\left(\frac xa\right)\right)'=\frac 1x\\ \left(-\log\left(\frac{1-x}b\right)\right)'=\frac 1{1-x}\\ \left(2x\right)'=2 $$ implying $$ h'(x)=\frac{\left(\frac1x+\frac1{1-x}\right)\cdot 2x-\left(\log\left(\frac{x}a\right)-\log\left(\frac{1-x}b\right)\right)\cdot 2}{(2x)^2} $$ And so finally, we can conclude that $g''(y)=\left(h(\sqrt{y})\right)'=h'(\sqrt y)/(2\sqrt y)$ which then is $$ g''(y)=\frac{\left(\frac1{\sqrt y}+\frac1{1-\sqrt y}\right)\cdot 2\sqrt y-\left(\log\left(\frac{\sqrt y}a\right)-\log\left(\frac{1-\sqrt y}b\right)\right)\cdot 2}{(2\sqrt y)^3} $$

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Finally, coming back to gerw's answer, to see whether $g$ is convex so that the inequality holds for $f$, we must check whether $g''$ is non-negative in $(0,1)$. In fact we do not have to check this for $g''$ but can work with $h'$ since those two share signs. Furthermore, only the numerator affects the sign of the expression. We can also remove the shared factor $2$ and simplify: $$ \left(\frac1x+\frac1{1-x}\right)\cdot x=\frac{1}{1-x} $$ and apply logarithmic rules to draw out $a,b$ and recombine. Then $h'(x)\geq 0$ can be seen to be equivalent to: $$ q(x)=\frac1{1-x}+\log(1-x)-\log(x)+\log(a/b) \geq 0 $$ By taking the derivative of $q$: $$ q'(x)=\frac 1{(1-x)^2}-\frac1{1-x}-\frac 1x=\frac{2x-1}{x(1-x)^2} $$ we see that $q$ has its minimum at $x=0.5$. One can check that $q(0.5)=2+\log(a/b)$ which implies that $q$ is non-negative as long as $$ \log(a/b)\geq -2\\ \Updownarrow\\ \frac ab\geq \operatorname e^{-2} $$

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Going back to the original problem in this question $a=b=1$ so in this case $g$ is convex and the original inequality works. In the other post you wanted: $$ a=p\\ b=1-p $$ so the requirement becomes $$ \frac p{1-p}\geq \operatorname e^{-2}\\ \Updownarrow\\ p\geq\frac{\operatorname e^{-2}}{1+\operatorname e^{-2}}=\frac{1}{\operatorname e^2+1}\approx 0.1192029220 $$ The last part is of course assuming $p>0$ for otherwise the inequality will be reversed, and/or the function $f$ will be ill-defined.

Answer 2

If you do a substitution, $y = x^2$, then \begin{equation*} \bar y = \bar x^2 = \frac{x_1^2+x_2^2}2 = \frac{y_1+y_2}2. \end{equation*} Thus, your question is equivalent to asking about convexity (w.r.t. $y$) of the function \begin{equation*} g(y) = \sqrt{y}\,\log\sqrt{y} + (1-\sqrt{y})\,\log(1-\sqrt{y}) . \end{equation*} This should be answered by taking second derivatives.

Similarly, you can attack your original question by looking at \begin{equation*} g_p(y) = \sqrt{y}\,\log\frac{\sqrt{y}}{p} + (1-\sqrt{y})\,\log\frac{1-\sqrt{y}}{1-p} . \end{equation*}

For a quick check, I used MATLAB and found that $g$ is convex, while $g_p$ fails to be convex for all $p$ smaller than something like $0.12$.