An enhanced Cauchy-Schwarz inequality of the integral form

We all know the very famous Cauchy-Schwaz inequality (of the integral form):

Let $f$ and $g$ be real functions continuous in $[a,b]$. Then the following equality holds:

$$\left(\int_a^bf(x)g(x)\mathrm{d}x\right)^2\le\int_a^b(f(x))^2\mathrm{d}x\int_a^b(g(x))^2\mathrm{d}x$$

The purpose of this blog is to prove an enhanced version of this inequality, which is stated as follows:

Let $f,g\in L^2[a,b]$, $0<m\le f\le M$, and $\int_a^bg(x)\mathrm{d}x=0$. Prove that

$$\left(\int_a^bf(x)g(x)\mathrm{d}x\right)^2\le\left(\frac{M-m}{M+m}\right)^2\int_a^b(f(x))^2\mathrm{d}x\int_a^b(g(x))^2\mathrm{d}x$$

Step 1: Substract $k$ from $f$

From the vanilla C-S inequality, we know that if $f$ and $g$ are linear, i.e., $f=kg$, the equality holds. However in this problem, we know that there does not exist such $k$ that $f=kg$ because $f>0$ and $g$ has non-positive values in $[a,b]$. This suggests us that, to reach its minimum value as much as possible, we should manage to push $f$ to be as linear as $g$ as possible. To this end, we can substract a constant $k$, from $f$, so that $f$ and $g$ can hold the same sign at any given input $x\in[a,b]$.

So what should we choose for $k$. An intuitive choice is the average of $f$, i.e., $k=\frac{1}{b-a}\int_a^bf(x)\mathrm{d}x$, which lies within the minimum value $m$ and maximum value $M$.

A more important property of substracting a constant number $k$ from $f$ is, it does not change the value of $\int_a^bf(x)g(x)\mathrm{d}x$ as $\int_a^bg(x)\mathrm{d}x=0$ is given. Then it follows that:

$$\left(\int_a^bf(x)g(x)\mathrm{d}x\right)^2=\left(\int_a^b\left(f(x)-k\right)g(x)\mathrm{d}x\right)^2\le \left[\int_a^bf(x)^2\mathrm{d}x-\frac{1}{b-a}\left(\int_a^bf(x)\mathrm{d}x\right)^2\right]\int_a^bg(x)^2\mathrm{d}x$$

The last inequality applies the standard C-S inequality.

Now all we need to do is to prove that:

$$\int_a^bf(x)^2\mathrm{d}x-\frac{1}{b-a}\left(\int_a^bf(x)\mathrm{d}x\right)^2\le \left(\frac{M-m}{M+m}\right)^2\int_a^bf(x)^2\mathrm{d}x$$

Step 2: Discriminant of a quadratic equation

Rearranging the above inequality, we have:

$$4Mm(b-a)\int_a^bf(x)^2\mathrm{d}x\le \left((M+m)\int_a^bf(x)\mathrm{d}x\right)^2$$

It is clear that if we treat $B=(M+m)\int_a^bf(x)\mathrm{d}x$, $A=\int_a^bf(x)^2\mathrm{d}x$ and $C=Mm(b-a)$, we obtain the discriminant of a quadratic equation:

$$B^2-4AC\ge 0$$

We can construct such quadratic equation:

$$p(y)=\left(\int_a^bf(x)^2\mathrm{d}x\right) y^2+\left((M+m)\int_a^bf(x)\mathrm{d}x\right)y + Mm(b-a)$$

Then $B^2-4AC$ is equivalent to fact that there exists at least one point on $p(y)$ where $p(y)\le 0$. The last of our work is to find such point.

Step 3: Find a point

The easiest way to find such point that the value of $p(y)$ on that point is non-positive is by trying some common values. For example, when $y=0$, $p(0)=Mm(b-a)>0$. When $y=1$, we have:

$$p(1)=\int_a^bf(x)^2\mathrm{d}x+(M+m)\int_a^bf(x)\mathrm{d}x + Mm(b-a)=\int_a^b\left(f(x)^2+\left(M+m\right)f(x)+Mm\right)\mathrm{d}x=\int_a^b(f(x)+M)(f(x)+m)\mathrm{d}x$$

It is a nice form but still not what we expect as the result is positive. However, if we change the sign of $B$, that is to turn $B$ to $-B$: $B=-(M+m)\int_a^bf(x)\mathrm{d}x$, the new $p(1)$ will become:

$$p(1)=\int_a^b(f(x)-M)(f(x)-m)\mathrm{d}x\le 0$$

This is exactly what we want: a value on $p(y)$ that is non-positive. Now $p(y)$ is:

$$p(y)=\left(\int_a^bf(x)^2\mathrm{d}x\right) y^2-\left((M+m)\int_a^bf(x)\mathrm{d}x\right)y + Mm(b-a)$$

and $p(1)\le 0$ implies the discriminant $B^2-4AC\ge 0$, which closes our proof.