Say you have a valid density function $f(x)$ from $[0, 1]$. Is there a function $f(x)$ such that $cf^2(x)$ is not a valid density function $\forall c \in \mathbb{R}$?
My understanding that a P.D.F. will be invalid if any of the following conditions are not met
- There is not an asymptote in its domain
- There are no negative values
- The integral equals 1.
If the first condition is met, then $\int_0^1 cf^2(x)$ will be some real value $b$, so chose $c=\frac{1}{b}$. Thus the third condition will be met.
Additionally an asymptote cannot be moved by squaring a function. Thus the first condition will never be an asymptote in the domain of $cf^2(x)$ if there is not one in $f(x)$. Which there cannot be, as $f(x)$ is a valid density function.
Finally, it is impossible to get a negative value from a square, therefore condition 2 will always be met for $cf^2(x)$.
Thus there is no such valid density function $f(x)$ where $cf^2(x)$ is not also a valid density function. However I know this is incorrect as this is from a question that states that there is such a function. Where am I wrong and what is a counter-example to this?