Is the function characterized by $f(\alpha x+(1-\alpha) y) \le f^{\alpha}(\alpha x)f^{1-\alpha}(y)$ convex?

Is a non-negative function $f(x)$ convex ? If for $x \ge y$ it satisfies for any $\alpha \in [0,1]$. \begin{align} f(\alpha x+(1-\alpha) y) \le f^{\alpha}(\alpha x)f^{1-\alpha}(y) \ \text{ eq.1} \end{align}

This is very reminiscent of the log-convexity which is defined as
\begin{align} f(\alpha x+(1-\alpha) y) \le f^{\alpha}( x)f^{1-\alpha}(y). \end{align}

1. There exist a log-convex function $g(x)$ such that $f(x) \le g(x)$ and such that \begin{align} f(\alpha x+(1-\alpha) y) \le g^\alpha(x)f^{1-\alpha}(y). \end{align}
2. $f(x)$ is a decreasing function of $x$.

A little back ground: I am trying to show that if $f$ satisfies eq.$1$ then it is continuous. The property that came to my mind is that if $f$ is convex then it is continuos on the open set. The condition in eq.$1$ is very similar to log-convexity (recall log-convexity implies convexity) and the hope is that it implies convexity.

Thank you for any help and suggestions, in advance.

[Note. This is an edited answer based on the helpful discussions with OP, as shown in the comments.]

Thanks for the interesting question. Here are some of my observations, and please let me know if I did something wrong. Thanks.

Observation 1

With $x = y$, we have from eq. 1, $$f(y) \le f^{\alpha}(\alpha y) \, f^{1-\alpha}(y).$$ Since $f(y)$ is nonnegative, $$f(y) \le f(\alpha y). \qquad (1)$$ This means the function $f(y)$ is decreasing for $y \ge 0$ and increasing for $y \le 0$.

A corollary is that $$f(x) \le f(0). \qquad (2)$$

Observation 2

Let $y < 0$ and $\alpha x = (1 - \alpha) (-y) > 0$, we have $$f(0) \le f^{\alpha}(\alpha x)\, f^{1-\alpha}(y).$$ But by (2) we have \begin{align} f(\alpha x) &\le f(0), \\ f(y) &\le f(0). \end{align} So $$f(0) \le f^{\alpha}(\alpha x)\, f^{1-\alpha}(y) \le f(0).$$ This is possible only if both $f(\alpha x)$ and $f(y)$ are equal to $f(0)$. In other words, $f(x)$ is a constant.

Observation 3

The above argument shows that eq. 1 is a very stringent condition on $f$, if $y$ can take any real number. However, if we impose the restriction $y \ge 0$, any monotonically decreasing function satisfies the requirement, eq. 1. This is because \begin{align} f(y) &\ge f(y + \alpha \, (x-y)), \\ f(\alpha x) &\ge f(\alpha x + (1 - \alpha) \, y), \end{align} So $$f^{1-\alpha}(y)f^\alpha(\alpha x) \ge f(\alpha x + (1- \alpha)y).$$

• Nice answer. Thank you very insightful. Do you see anyway of dropping the assumption of moronically decreasing function? For example, can we do anything with extra hypothesis 1. – Boby Oct 30 '15 at 20:56
• $g(x)$ is log-convex. I will correct the questions. Thanks – Boby Oct 30 '15 at 21:13
• @Boby. Thanks, I updated the answer. It seems that eq. 1 alone implies that $f(x)$ is a constant, if I were correct in observation 2. Do you see any mistakes in the argument? Or should we replace eq. 1 with something less stringent? Thanks. – hbp Oct 30 '15 at 22:01
• No, I don't see any wrong with you reasoning. Great intuition. Now that I started thinking about I actually wanted to study: $f( \alpha x+(1-\alpha) y) \le f^\alpha \left( \frac{x}{\alpha} \right) f^{1-\alpha}(y)$. I think this a more interesting question. What do you think.? – Boby Oct 30 '15 at 22:30
• @Boby. Second thought. I think I can use the bounty to fund the follow up question, how about that? Thank you very much for your generosity! – hbp Nov 4 '15 at 16:59