# integrability

#### rondo09

let f and g be integrable functions on [a,b]. Show that

$$\displaystyle |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^{2})^{1/2} {({\int_{a}^{b}} g^{2})^{1/2}$$

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#### tonio

let f and g be integrable functions on [a,b]. Show that

$$\displaystyle |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^{1/2})^2 {({\int_{a}^{b}} g^{1/2})^2$$

First, there's a (rather huge and important) mistake: it must be $$\displaystyle |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^2)^{1/2} {({\int_{a}^{b}} g^2)^{1/2}$$.

Second, you can google "Cauchy-Schwartz inequality" , or: let $$\displaystyle \lambda\in\mathbb{R}$$ , and then

$$\displaystyle 0\leq\int\left(f-\lambda g\right)^2=\int\left(f^2-2fg\lambda+g^2\lambda^2\right)$$ , and now take $$\displaystyle \lambda:=\frac{\int f^2}{\int fg}$$ , so substituting above:

$$\displaystyle 0\leq \int \!\!f^2 -\frac{\int f^2}{\int fg}\,2\int \!\!fg+\frac{\left(\int f^2\right)^2}{\left(\int fg\right)^2}\,\int \!\!g^2$$ $$\displaystyle =-\int \!\!f^2+\frac{\left(\int f^2\right)^2}{\left(\int fg\right)^2}\,\int \!\!g^2\iff$$

$$\displaystyle \iff \left(\int\!fg\right)^2\int\!f^2\leq \left(\int\!f^2\right)^2\int\!g^2$$

And from here your inequality follows at once ( note that we must assume something to justify the above proof: what is this and what happens if the assumption isn't fulfilled?)

Tonio

#### rondo09

First, there's a (rather huge and important) mistake: it must be $$\displaystyle |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^2)^{1/2} {({\int_{a}^{b}} g^2)^{1/2}$$.

Second, you can google "Cauchy-Schwartz inequality" , or: let $$\displaystyle \lambda\in\mathbb{R}$$ , and then

$$\displaystyle 0\leq\int\left(f-\lambda g\right)^2=\int\left(f^2-2fg\lambda+g^2\lambda^2\right)$$ , and now take $$\displaystyle \lambda:=\frac{\int f^2}{\int fg}$$ , so substituting above:

$$\displaystyle 0\leq \int \!\!f^2 -\frac{\int f^2}{\int fg}\,2\int \!\!fg+\frac{\left(\int f^2\right)^2}{\left(\int fg\right)^2}\,\int \!\!g^2$$ $$\displaystyle =-\int \!\!f^2+\frac{\left(\int f^2\right)^2}{\left(\int fg\right)^2}\,\int \!\!g^2\iff$$

$$\displaystyle \iff \left(\int\!fg\right)^2\int\!f^2\leq \left(\int\!f^2\right)^2\int\!g^2$$

And from here your inequality follows at once ( note that we must assume something to justify the above proof: what is this and what happens if the assumption isn't fulfilled?)

Tonio
Yeah.. I have mistyped it.. Sorry... It should be
$$\displaystyle |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^2)^{1/2} {({\int_{a}^{b}} g^2)^{1/2}$$

#### Also sprach Zarathustra

You can prove it also similarly to what tonio did and using using ax^2+bx+c>=0 iff b^2-4ac<0