[Edit. The question lacks certain important conditions, as kindly pointed out by NeutralElement. Below is the amended version. I apologize for the omissions and many thanks to NeutralElement and user254665 for helpful comments. ]

This is related to the question raised by Boby. For a variant, consider the nonnegative function $f(x)$ satisfied by $$ f(\alpha\, x + (1 - \alpha) \, y) \le f^{\alpha}(x/\alpha) \, f^{1-\alpha}(y), \qquad (1) $$ for $\alpha \in (0, 1]$ and real $x$ and $y$.

What can we say about his function? Is it always convex?

[Edit2. In the original question by Boby, there is one additional condition that requires that $x \ge y$. But unfortunately I missed this condition in the previous edit. So I'll settle for the result for the above question. I apologize for the many omissions. However, if anyone can comment how this condition changes the result, it would be much appreciated. Thank you.]

Here are some observations that may or may not help.

Observation 1

With $y = x$, we have $$ f(x) \le f(x/\alpha). \qquad (2) $$ This means that $f(x)$ is increasing for $x \ge 0$, and decreasing for $x \le 0$. A corollary is that $f(0)$ is the global minimum, i.e, $$ f(0) \le f(x). \qquad (3) $$

Observation 2

By Young's inequality or the weighted AM-GM inequality, we have $$ f^\alpha(x/\alpha) f^{1-\alpha}(y) \le \alpha f(x/\alpha) + (1 - \alpha) f(y). \qquad (4) $$ Thus, (4) and (1) require $$ f(y + \alpha (x - y)) \le \alpha \, f(x/\alpha) + (1 - \alpha) \, f(y). $$

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    $\begingroup$ your statement lacks essential details. Is it for every $\alpha\in(0,1]$ and every $x,y\in\mathbb{R}$? Please amend your question. $\endgroup$ – Neutral Element Nov 9 '15 at 14:12
  • $\begingroup$ @NeutralElement. Thank you again for pointing out that $f(x)$ should be nonnegative, and the domain of $\alpha$, $x$, and $y$. I sincerely apologize for the mistakes. $\endgroup$ – hbp Nov 9 '15 at 19:23
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    $\begingroup$ I take it that the second last eq. ,which is unlabelled, is "(4)"? Then "(4) and (1) require..." $\endgroup$ – DanielWainfleet Nov 10 '15 at 4:59
  • $\begingroup$ @user254665. Thank you for pointing this out. Corrected. $\endgroup$ – hbp Nov 10 '15 at 8:05

Some ideas below.

Since fractional powers are defined only for non-negative numbers this problem should have in its statement $f\ge0$ which comes to $f(0)\ge0$ since $0$ is the global min.

If $f(0)=0$ then pick $y=0$ in the given condition to get $f(\alpha x)\le0$ for every $\alpha\in(0,1],\ x\in\mathbb{R}$. In this case $f\equiv0$.

It remains that $f(0)>0$. Clearly $f=k>0$ constant is a solution for this problem.

Let us look for non-constant solutions. Again for $y=0$ we get $$ f(\alpha x)\le f^\alpha(x/\alpha)f^{1-\alpha}(0) $$

Assume that $f$ is non-constant. Then $\lim_{x\to+\infty}f(x)=\sup_{x\ge0} f(x)=+\infty$. Similarly, $\lim_{x\to-\infty}f(x)=\sup_{x\le0} f(x)=+\infty$.

More precisely, if we assume that $\lim_{x\to+\infty}f(x)=\ell=\sup_{x\ge0} f(x)<+\infty$ then $\ell\ge f(0)>0$. Pick $x=\alpha^{-2}$ and let $\alpha\to 0$ to get $\ell\le f(0)$ from which we obtain the contradiction $f\equiv f(0)$.

Now I see that a sharper inequality is obtained for $x=0$. Namely, $$ f(\alpha x)\le f^\alpha(x)f^{1-\alpha}(0) $$ (I renamed $y\to x,\ 1-\alpha\to\alpha$ )

Also I see that $f(x):=e^{|x|}$ is a solution so I will be concerned in the sequel only with the convexity of $f$.

Something else. Note that $f(x)=e^{g(x)}$ is a solution of this problem iff for every $\alpha\in(0,1],\ x,y\in\mathbb{R}$, $g(\alpha x+(1-\alpha)y)\le\alpha g(x/\alpha)+(1-\alpha)g(y)$.

This remark shows that your problem is equivalent to asking whether $g$ is convex whenever $g(\alpha x+(1-\alpha)y)\le\alpha g(x/\alpha)+(1-\alpha)g(y)$ holds for every $\alpha\in(0,1],\ x,y\in\mathbb{R}$ (and that is a simplification of your problem).

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    $\begingroup$ Thanks for answering, but I am running out of votes today, will up vote your answer later :-) $\endgroup$ – hbp Nov 9 '15 at 19:56
  • $\begingroup$ The proof of $f(+\infty) = +\infty$ and simplification are quite nice. Thank you again for your beautiful answer. Unfortunately, I just realized missed another condition $x \ge y$ when copying Boby's question. Will change the result a bit? Anyway, I'm happy to see the result either with and without $x \ge y$. Thanks again. $\endgroup$ – hbp Nov 10 '15 at 4:07
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    $\begingroup$ Actually the condition $x\ge y$ might help. Please restate the problem with all the details you wanted. Stop thanking and apologizing. It's just Math (a hobby). Read the rules of the forum. $\endgroup$ – Neutral Element Nov 10 '15 at 20:46

Statement of the problem: What can we say about a function $f:\mathbb{R}\to\mathbb{R}$ with the property that $$ (P)\ \ \forall \alpha\in(0,1],\ x,y\in\mathbb{R},\ x\ge y,\ \ f(\alpha x+(1-\alpha)y)\le f^{\alpha}(x/\alpha)f^{1-\alpha}(y)? $$

Since we can take $x=y$, $f(0)$ is still the global minimum value.

Assume that $f(0)=0$. We claim in this case that $f\equiv 0$. Indeed, take $y=0$ in (P) to get $\forall \alpha\in(0,1],\ x\ge 0,\ \ f(\alpha x)\le 0$ from which $f(x)=0$ for every $x\ge0$. Similarly, make $x=0$ in (P) to get $\forall \alpha\in(0,1],\ y\le 0,\ \ f((1-\alpha) y)\le 0$ from which $f(y)=0$ for every $y\le0$.

It remains that $f(0)>0$. Clearly $f=k>0$ constant is a solution for this problem.

We claim that if $f$ is non-constant on $\mathbb{R}_{+}$ then $\lim_{x\to+\infty}f(x)=+\infty$.

Again for $y=0$ we get $$ \forall \alpha\in(0,1],\ x\ge 0,\ \ f(\alpha x)\le f^\alpha(x/\alpha)f^{1-\alpha}(0) $$

If we assume that $\lim_{x\to+\infty}f(x)=\ell=\sup_{x\ge0} f(x)<+\infty$ then $\ell\ge f(0)>0$. Pick $x=\alpha^{-2}$ to get $$ \forall \alpha\in(0,1],\ \ f(\alpha^{-1})\le f^\alpha(\alpha^{-3})f^{1-\alpha}(0)\le\ell^{\alpha} f^{1-\alpha}(0) $$

and pass to supremum over $\alpha$ to get $\ell\le f(0)$ from which we get $f\equiv f(0)$.

Similarly, if $f$ is non-constant on $\mathbb{R}_{-}$ then $\lim_{x\to-\infty}f(x)=+\infty$. (This time pick $x=0$.)

However, let us apply the natural logarithm to the inequality to get for $g(x)=\ln f(x)$ that $$ (PL)\ \ \ \forall \alpha\in(0,1],\ x,y\in\mathbb{R},\ x\ge y,\ \ g(\alpha x+(1-\alpha)y)\le {\alpha}g(x/\alpha)+(1-\alpha)g(y) $$

We have $f$ has (P) iff $g$ has (PL); both $f$ and $g$ satisfy (PL); and we are looking for non-constant solutions.

Again from $x=y$ we know that $g(0)$ is the global min value, $g$ is decreasing on the negative $x-$axis and increasing on the positive $x-$axis, and $\lim_{x\to\pm\infty}g(x)=+\infty$. We can assume that $g$ is non-negative with $g(0)=0$ otherwise we replace it by $g-g(0)$.

Because $0$ is a global minimum we have: $g$ is convex on $\mathbb{R}$ iff $g$ is convex on $\mathbb{R}_{-}$ and $g$ is convex on $\mathbb{R}_{+}$.

Consider the function $g(x)=\sqrt{x}$, for $x\ge0$, $g(x)=0$, for $x<0$.

Then for $\alpha\in(0,1)$, $x\ge y$ we consider the cases:

(A) $\alpha x+(1-\alpha)y\le0$,

(B) $\alpha x+(1-\alpha)y>0$, $x\ge0\ge y$ and

(C) $\alpha x+(1-\alpha)y>0$, $x\ge y\ge0$.

Condition (PL) is easily verified in case (A) because $g(\alpha x+(1-\alpha)y)=0$ and $g\ge0$.

In case (B) we have $g(\alpha x+(1-\alpha)y)\le\alpha g(x/\alpha)+(1-\alpha)g(y)\Leftrightarrow\sqrt{\alpha x+(1-\alpha)y}\le\alpha\sqrt{x/\alpha}=\sqrt{\alpha x}$ which is true because $\alpha x+(1-\alpha)y>0$ and $x\ge0\ge y$.

In case (C) $$ g(\alpha x+(1-\alpha)y)\le\alpha g(x/\alpha)+(1-\alpha)g(y)\Leftrightarrow $$ $$ \sqrt{\alpha x+(1-\alpha)y}\le\alpha\sqrt{x/\alpha}+(1-\alpha)\sqrt{y}=\sqrt{\alpha x}+(1-\alpha)\sqrt{y}\Leftrightarrow $$ $$ \alpha x+(1-\alpha)y\le\alpha x+(1-\alpha)^{2}y+2(1-\alpha)\sqrt{\alpha xy}\Leftrightarrow\alpha y\le2\sqrt{\alpha xy} $$ which is true because $\alpha\le\sqrt{\alpha}$ and $y\le\sqrt{xy}$.

Hence $g$ satisfies (PL) and is not convex (on $\mathbb{R}$). Therefore $f(x)=e^{\sqrt{x}}$, for $x\ge0$, $f(x)=1,$ for $x<0$ is continuos, satisfies (P) but is not convex (it is concave on $[0,1]$). So your modified question has a negative answer.

  • $\begingroup$ This is a very nice answer. Thank you! $\endgroup$ – hbp Nov 11 '15 at 7:50

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