# minkowski inequality

• Jan 2nd 2010, 03:00 PM
Archi
minkowski inequality
Let $\displaystyle (X,F,\mu)=([0,a],\beta, \Lambda)$ be the usual Lebesgue measurable space.
Using Holder inequality, prove Minkowski inequality for any $\displaystyle f \in L^2(X),g \in L^2(X)$, that is $\displaystyle ||f+g||_2 \leq ||f||_2 +||g||_2$

i am stuck and dont know what to do. any help would be appreciated.
• Jan 3rd 2010, 07:20 AM
putnam120
Imitate the proof of the triangle inequality. (that is start by squaring both sides) Then you can use Holder's inequality with $\displaystyle p=q=2$
• Jan 4th 2010, 03:09 PM
Archi
Quote:

Originally Posted by putnam120
Imitate the proof of the triangle inequality. (that is start by squaring both sides) Then you can use Holder's inequality with $\displaystyle p=q=2$

i am not sure what i should square the both sides of. holders inequality?
• Jan 4th 2010, 07:23 PM
putnam120
$\displaystyle \parallel f+g\parallel_2^2=\int (f+g)^2=\int f^2+\int g^2+2\int fg$

$\displaystyle (\parallel f\parallel_2 +\parallel g\parallel_2)^2=\int f^2+\int g^2+2\parallel f\parallel_2\parallel g\parallel_2$

From here you should be able to see how to use Holder's.