integrability

Aug 2010
13
0
let f and g be integrable functions on [a,b]. Show that

\(\displaystyle |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^{2})^{1/2} {({\int_{a}^{b}} g^{2})^{1/2}\)
 
Last edited:
Oct 2009
4,261
1,836
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
 
Aug 2010
13
0
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
\(\displaystyle |\int_{a}^{b} fg| \leq {({\int_{a}^{b}} f^2)^{1/2} {({\int_{a}^{b}} g^2)^{1/2}\)
 
Dec 2009
1,506
434
Russia
You can prove it also similarly to what tonio did and using using ax^2+bx+c>=0 iff b^2-4ac<0