The conditions on $F$ are

1. $F(0)=0$

2. $F(\overline{\alpha}) = 1$

3. $F'(x) \gt 0$ for all $x \in (0, \overline{\alpha})$

4. $\int_0^{\overline{\alpha}} x F'(x) dx = 1$.

5. $F'$ is differentiable.

These are all nonlocal in the sense that sufficiently small (smooth) perturbations of $F$ within small neighborhoods of any finite discrete set of points in $(0, \overline{\alpha})$ can be found which preserve them all. By applying such a local perturbation we can modify $F'$ within a narrow interval to be as large as we want without changing any of the conditions.

The figures below may help in following this argument.

I claim there exists at least one $x_0 \in (0, \overline{\alpha})$ for which $(N-1)(1-\delta+\delta F(x_0)) - \delta x_0 F'(x_0) \gt 0$. For if not, a simple comparison shows that $F(0)$ must be less than the value at $0$ of the (unique) solution to the first-order ordinary differential equation for $F$,

$$L[F](x) = (N-1)(1-\delta+\delta F(x)) - \delta x F'(x) = 0, \quad F(\overline{\alpha}) = 1.$$

This equation can explicitly be solved and its value at $0$ found to equal $1 - 1/\delta \lt 0$, whence $F(0) \lt 0$, violating the first condition.

Let's perturb $F$ to $\widehat{F}$. The following argument needs only routine constructions to be made rigorous: mix in tiny amounts of a distribution supported in an arbitrarily small neighborhood of $x_0$ and another distribution supported in a neighborhood of an $x_1$ on the other side of $1$ from $x_0$ in such a way that the expectation of $F$ remains unchanged. By making the first of these mixed-in distributions have sufficiently small support, we can cause it to increase $F'$ at $x_0$ by any desired amount while changing $F(x_0)$ arbitrarily little. In this fashion we can cause $L[\widehat{F}](x_0)$ to range from $L[F](x_0)$ down to $-\infty$ in a continuous manner. Do this so that $L[\widehat{F}](x_0) = 0$.

Therefore, there exist no restrictions whatsoever on any of the moments of $F$. That explains why you had such difficulty obtaining any!

In these figures, the plots associated with the original $F$ is shown in blue and those associated with the perturbed CDF $\widehat{F}$ are shown in red. As $\varepsilon$ shrinks, the two spikes in the PDF and $L[F]$ grow longer (vertically). Here, I noticed that $L[F](3/2) \gt 0$, and so perturbed $F$ near $3/2$ and near a counterbalancing value $x_1 = 1/2$. This creates two tiny apparent jumps around $3/2$ and $1/2$ in $F$, but upon closer inspection they are smooth--just steep. Their steepness makes $L[F]$ small. Because the jumps can be made arbitrarily steep, the spikes in $L[F]$ can be extended below zero, creating a zero-crossing: that solves the problem. (Just for fun I have used an $F$ that fails to be twice differentiable at $0$: $F'$ diverges there. The construction still goes through.)

The conditions on $F$ are

1. $F(0)=0$

2. $F(\overline{\alpha}) = 1$

3. $F'(x) \gt 0$ for all $x \in (0, \overline{\alpha})$

4. $\int_0^{\overline{\alpha}} x F'(x) dx = 1$.

5. $F'$ is differentiable.

These are all nonlocal in the sense that sufficiently small (smooth) perturbations of $F$ within small neighborhoods of any finite discrete set of points in $(0, \overline{\alpha})$ can be found which preserve them all. By applying such a local perturbation we can modify $F'$ within a narrow interval to be as large as we want without changing any of the conditions.

The figures below may help in following this argument.

I claim there exists at least one $x_0 \in (0, \overline{\alpha})$ for which $(N-1)(1-\delta+\delta F(x_0)) - \delta x_0 F'(x_0) \gt 0$. For if not, a simple comparison shows that $F(0)$ must be less than the value at $0$ of the (unique) solution to the first-order ordinary differential equation for $F$,

$$L[F](x) = (N-1)(1-\delta+\delta F(x)) - \delta x F'(x) = 0, \quad F(\overline{\alpha}) = 1.$$

This equation can explicitly be solved and its value at $0$ found to equal $1 - 1/\delta \lt 0$, whence $F(0) \lt 0$, violating the first condition.

Let's perturb $F$ to $\widehat{F}$. The following argument needs only routine constructions to be made rigorous: mix in tiny amounts of a distribution supported in an arbitrarily small neighborhood of $x_0$ and another distribution supported in a neighborhood of an $x_1$ on the other side of $1$ from $x_0$ in such a way that the expectation of $F$ remains unchanged. By making the first of these mixed-in distributions have sufficiently small support, we can cause it to increase $F'$ at $x_0$ by any desired amount while changing $F(x_0)$ arbitrarily little. In this fashion we can cause $L[\widehat{F}](x_0)$ to range from $L[F](x_0)$ down to $-\infty$ in a continuous manner. Do this so that $L[\widehat{F}](x_0) = 0$.

Therefore, there exist no restrictions whatsoever on any of the moments of $F$. That explains why you had such difficulty obtaining any!

In these figures, the plots associated with the original $F$ are shown in blue and those associated with the perturbed CDF $\widehat{F}$ are shown in red. As $\varepsilon$ shrinks, the two spikes in the PDF and $L[F]$ grow longer (vertically). Here, I noticed that $L[F](3/2) \gt 0$, and so perturbed $F$ near $3/2$ and near a counterbalancing value $x_1 = 1/2$. This creates two tiny apparent jumps around $3/2$ and $1/2$ in $F$, but upon closer inspection they are smooth--just steep. Their steepness makes $L[F]$ small. Because the jumps can be made arbitrarily steep, the spikes in $L[F]$ can be extended below zero, creating a zero-crossing: that solves the problem. (Just for fun I have used an $F$ that fails to be twice differentiable at $0$: $F'$ diverges there. The construction still goes through.)

The point is that we can always create such spikes (and give them arbitrarily small area, thereby changing $F$ by arbitrarily small amounts), so to obtain a zero-crossing it's enough to show that $L[F]$ must have some neighborhood in which it is positive. But if it is not, $F$ will fail to be a CDF: consistently negative values of $L[F]$ mean that $F'$ is too large, on average, and so if $F$ ends up with a limiting value of $1$ at the right--as it must in order to be a CDF--also it must have a negative value at $0$, which is not allowable.

The conditions on $F$ are

1. $F(0)=0$

2. $F(\overline{\alpha}) = 1$

3. $F'(x) \gt 0$ for all $x \in (0, \overline{\alpha})$

4. $\int_0^{\overline{\alpha}} x F'(x) dx = 1$.

5. $F'$ is differentiable.

These are all nonlocal in the sense that sufficiently small (smooth) perturbations of $F$ within small neighborhoods of any finite discrete set of points in $(0, \overline{\alpha})$ can be found which preserve them all. By applying such a local perturbation we can modify $F'$ within a narrow interval to be as large as we want without changing any of the conditions.

I claim there exists at least one $x_0 \in (0, \overline{\alpha})$ for which $(N-1)(1-\delta+\delta F(x_0)) - \delta x_0 F'(x_0) \gt 0$. For if not, a simple comparison shows that $F(0)$ must be less than the value at $0$ of the (unique) solution to the first-order ordinary differential equation for $F$,

$$L[F](x) = (N-1)(1-\delta+\delta F(x)) - \delta x F'(x) = 0, \quad F(\overline{\alpha}) = 1.$$

This equation can explicitly be solved and its value at $0$ found to equal $1 - 1/\delta \lt 0$, whence $F(0) \lt 0$, violating the first condition.

Let's perturb $F$ to $\widehat{F}$. The following argument needs only routine constructions to be made rigorous: mix in tiny amounts of a distribution supported in an arbitrarily small neighborhood of $x_0$ and another distribution supported in a neighborhood of an $x_1$ on the other side of $1$ from $x_0$ in such a way that the expectation of $F$ remains unchanged. By making the first of these mixed-in distributions have sufficiently small support, we can cause it to increase $F'$ at $x_0$ by any desired amount while changing $F(x_0)$ arbitrarily little. In this fashion we can cause $L[\widehat{F}](x_0)$ to range from $L[F](x_0)$ down to $-\infty$ in a continuous manner. Do this so that $L[\widehat{F}](x_0) = 0$.

Therefore, there exist no restrictions whatsoever on any of the moments of $F$. That explains why you had such difficulty obtaining any!

The conditions on $F$ are

1. $F(0)=0$

2. $F(\overline{\alpha}) = 1$

3. $F'(x) \gt 0$ for all $x \in (0, \overline{\alpha})$

4. $\int_0^{\overline{\alpha}} x F'(x) dx = 1$.

5. $F'$ is differentiable.

These are all nonlocal in the sense that sufficiently small (smooth) perturbations of $F$ within small neighborhoods of any finite discrete set of points in $(0, \overline{\alpha})$ can be found which preserve them all. By applying such a local perturbation we can modify $F'$ within a narrow interval to be as large as we want without changing any of the conditions.

The figures below may help in following this argument.

I claim there exists at least one $x_0 \in (0, \overline{\alpha})$ for which $(N-1)(1-\delta+\delta F(x_0)) - \delta x_0 F'(x_0) \gt 0$. For if not, a simple comparison shows that $F(0)$ must be less than the value at $0$ of the (unique) solution to the first-order ordinary differential equation for $F$,

$$L[F](x) = (N-1)(1-\delta+\delta F(x)) - \delta x F'(x) = 0, \quad F(\overline{\alpha}) = 1.$$

This equation can explicitly be solved and its value at $0$ found to equal $1 - 1/\delta \lt 0$, whence $F(0) \lt 0$, violating the first condition.

Let's perturb $F$ to $\widehat{F}$. The following argument needs only routine constructions to be made rigorous: mix in tiny amounts of a distribution supported in an arbitrarily small neighborhood of $x_0$ and another distribution supported in a neighborhood of an $x_1$ on the other side of $1$ from $x_0$ in such a way that the expectation of $F$ remains unchanged. By making the first of these mixed-in distributions have sufficiently small support, we can cause it to increase $F'$ at $x_0$ by any desired amount while changing $F(x_0)$ arbitrarily little. In this fashion we can cause $L[\widehat{F}](x_0)$ to range from $L[F](x_0)$ down to $-\infty$ in a continuous manner. Do this so that $L[\widehat{F}](x_0) = 0$.

Therefore, there exist no restrictions whatsoever on any of the moments of $F$. That explains why you had such difficulty obtaining any!

In these figures, the plots associated with the original $F$ is shown in blue and those associated with the perturbed CDF $\widehat{F}$ are shown in red. As $\varepsilon$ shrinks, the two spikes in the PDF and $L[F]$ grow longer (vertically). Here, I noticed that $L[F](3/2) \gt 0$, and so perturbed $F$ near $3/2$ and near a counterbalancing value $x_1 = 1/2$. This creates two tiny apparent jumps around $3/2$ and $1/2$ in $F$, but upon closer inspection they are smooth--just steep. Their steepness makes $L[F]$ small. Because the jumps can be made arbitrarily steep, the spikes in $L[F]$ can be extended below zero, creating a zero-crossing: that solves the problem. (Just for fun I have used an $F$ that fails to be twice differentiable at $0$: $F'$ diverges there. The construction still goes through.)