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This is from a Real Analysis class:

For $f,g \in R[a,b],$ show that $\left|\int_a^b f g \right| \le \sqrt{\int_a^b f^2 \int_a^b g^2}$

I was given the hint to expand $\int_a^b (xf+g)^2$ to a quadratic in $x$ and use its discriminant. Obviously the discriminant $D=0$ here, but I have no idea how this helps. Can anyone enlighten me on this?

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  • $\begingroup$ No, the discriminant $D$ is not zero. The discriminant of $(xf+g)^2$ is $0$. But that wasn't what was asked. $\endgroup$ Apr 24, 2014 at 3:34
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    $\begingroup$ It's also a bit confusing to use $x$ here, because $x$ might be the variable of integration. In the hint, assume it is not the variable of integration, that is, find the quadratic function $$p(x)=\int_a^b (xf(t)+g(t))^2\,dt$$ $\endgroup$ Apr 24, 2014 at 3:37
  • $\begingroup$ It is worth mentioning that this is a special case of Cauchy-Schwarz inequality (a.k.a. Cauchy-Bunyakovski inequality). You can probably find several posts about this inequality of this site - for example I was able to find this one with a little searching around: math.stackexchange.com/questions/183855/… $\endgroup$ Apr 24, 2014 at 5:28

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Note that $$\begin{align}p(x)&=\int_a^b (xf(t)+g(t))^2\,dt\\ &=A x^2+ Bx+C \end{align}$$ Where: $$\begin{align} A&=\int_a^b f^2\\ B&=2\int_a^b fg\\ C&=\int_a^b g^2 \end{align}$$

Since $p(x)$ cannot be negative (why?), the discriminant of $p(x)$ cannot be positive (why?)

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  • $\begingroup$ Okay, so $p(x)$ can't be negative because it's the integral of a square function, and its discriminant can't be positive because there are no real zeros of an always-positive function... I think I can figure it out from here. Thanks a lot! $\endgroup$ Apr 24, 2014 at 4:12
  • $\begingroup$ I guess it could still have one zero, but that's what the $\le$ is for $\endgroup$ Apr 24, 2014 at 4:19
  • $\begingroup$ Yes. Basically, if there is an $x$ such that $xf(t)+g(t)=0$ for all $t$, there can be one zero. $\endgroup$ Apr 24, 2014 at 4:22

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