probability distribution with maximal standard deviation Suppose $f(x)$ is some probability distribution for a real random variable $x$
defined over $[0,1]$. Assume that the the function is continuous and differentiable.
Is there a a function $f$ for which the standard deviation of
$x$ is maximum? (I tried with calculus of variation but it seems that approach is not
working.)
 A: No.
We will solve the same problem but instead require that $X$ take values over the interval $[-\frac{1}{2}, \frac{1}{2}]$. This will make the math easier. To do this formally, define $X = x - 1/2$; the two will have the same variance. I'll also discuss maximising variance rather than standard deviation, but obviously it's an equivalent task.
Note that $Var(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2$. Note that $\mathbb{E}[X^2] \leq \frac{1}{4}$, with equality holding iff $P(X = \pm \frac{1}{2}) = 1$. And note that $\mathbb{E}[X]^2 \geq 0$, with equality holding iff $\mathbb{E}[X] = 0$. Therefore, the unique probability distribution for $X$ that maximises the variance is the one where $X$ takes the value $-\frac{1}{2}$ half the time and $\frac{1}{2}$ the other half, since this is the only distribution where $X$ is always equal to $\pm \frac{1}{2}$ and $\mathbb{E}[X] = 0$.
Note that obviously, this unique probability distribution which maximises $Var(X)$ is not continuous. Thus, we see that any continuously differentiable probability distribution will result in a variance strictly less than $\frac{1}{4}$.
However, we can come arbitrarily close to a variance of $\frac{1}{4}$ using very simple probability distribution functions of the form $f_n(x) = (2n + 1) (2x)^n$ for $n$ an integer. We see that the expected value of a variable with this PDF is clearly $0$, since it is symmetric about the origin. And we have $\mathbb{E}[X^2] = \int\limits_{-1/2}^{1/2} (2n + 1) x^2 (2x)^n dx = \frac{2n + 1}{4 (2n + 3)}$. And this ratio approaches $\frac{1}{4}$ as $n$ grows arbitrarily large.
Since we can get arbitrarily close to a variance of $\frac{1}{4}$ with a continuously differentiable PDF but cannot actually reach it, we see that there is no continuously differentiable $f$ which maximises the variance.
