I have the following question from a past analysis qualifying exam:
Let $f$ be a real-valued differentiable monotone function on $[0,1]$. Define $$g(x)=\frac12[f(x)+f(1-x)]$$ for $0\le x\le1$. Show that $$\sigma^2(g)\le\frac12\sigma^2(f)$$ where $$\sigma^2(u)=\int_0^1(u(x)-E(u))^2\;dx$$ and $E(u)=\int_0^1u(x)\;dx.$
I'm not really sure where to begin so I wrote out $\sigma^2(g)$ to try to get an idea of what to do here. $$\sigma^2(g)=\int_0^1\left(g(x)-\int_0^1g(t)\;dt\right)^2\;dx$$ $$=\int_0^1\left[\frac12(f(x)+f(1-x))-\frac12\int_0^1(f(t)+f(1-t))\;dt\right]^2\;dx$$ $$=\frac14\left(\int_0^1\left[f(x)+f(1-x)-\int_0^1(f(t)+f(1-t))\;dt\right]^2\;dx\right)$$ $$=\frac14\left(\int_0^1\left[f(x)-\int_0^1f(t)\;dt+f(1-x)-\int_0^1f(1-t)\;dt\right]^2\;dx\right)$$ $$=\frac14\left(\int_0^1\left[f(x)-\int_0^1f(t)\;dt\right]^2\;dx+\int_0^1\left[f(1-x)-\int_0^1f(1-t)\;dt\right]^2\;dx-2\int_0^1\left(f(x)-\int_0^1f(t)\;dt\right)\left(f(1-x)-\int_0^1f(1-t)\;dt\right)\;dx\right)$$
I've also noticed that by a change of variables, $\int_0^1f(1-t)\;dt=\int_0^1f(t)\;dt$, but I'm not sure if this will be useful. Additionally, I haven't used the fact that $f$ is differentiable or monotone yet.
Perhaps there is some way to rewrite the integrand so that Cauchy-Schwarz can be applied? Any help would be appreciated.
We may assume without loss of generality that $f$ has mean zero. Then we just have to prove that
$$\sigma^2(g)=\frac{1}{4}\int_{0}^{1}f(x)^2+f(1-x)^2+2 f(x)f(1-x)\,dx \leq \frac{1}{2}\int_{0}^{1}f(x)^2\,dx =\sigma^2(f)$$ i.e. $$ \int_{0}^{1} f(x) f(1-x)\,dx \leq 0. $$ Let $F(x)=\int_{0}^{x}f(t)\,dt$. Our assumption gives $F(0)=F(1)=0$, and as the antiderivative of a decreasing and differentiable function, $F$ is concave. By integration by parts $$\begin{eqnarray*} \int_{0}^{1}f(x)f(1-x)\,dx &=& \left[F(x)f(1-x)\right]_{0}^{1}+\int_{0}^{1}F(x)f'(1-x)\,dx\\ &=& \int_{0}^{1}F(x)f'(1-x)\,dx\end{eqnarray*} $$ where the concavity of $F(x)$ ensures $F(x)\geq 0$ and the monotonicity of $f(x)$ ensures $f'(1-x)\leq 0$.