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}
Extra Hypothesis we can add: 


*

*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}

*$f(x)$ 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.
 A: [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
$$
f^{1-\alpha}(y)f^\alpha(\alpha x) \ge f(\alpha x + (1- \alpha)y).
$$
