I have convinced myself that $$\int_0^1f(x)\cdot x^{n+1}\text{d}x > \int_0^1f(x)\cdot x^n\text{d}x \cdot \int_0^1f(x)\cdot x\text{d}x$$ is true whenever

  • $f$ is non-negative,
  • $\int_0^1f(x)\text{d}x=1$, and
  • it is not the case that $f(x)=\delta(x-c)$, where $\delta$ is the Dirac delta function and $c$ is a constant in $[0,1]$ (in which case it is obvious that equality holds instead).

However, I could use some help proving it. I would be very happy just to get a hint - no full solution needed.

Btw, if there's someone who think they can help but don't know what the Dirac delta function is, then just assume that $f$ is a normal function and ignore the last condition.


Presumably $n > 0$ here.

This is essentially a special case of the Harris inequality for the probability measure $f(x)\; dx$ on $[0,1]$. For any strictly increasing functions $g$ and $h$ we have

$$ \int_0^1 f(x) g(x) h(x)\; dx > \int_0^1 f(x) g(x)\; dx \cdot \int_0^1 f(x) h(x)\; dx $$ which follows from $$ \int_0^1 \int_0^1 (g(x)-g(y))(h(x)-h(y)) f(x) f(y)\; dx \; dy > 0$$


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