# A very annoying Inequality

• Nov 9th 2008, 10:58 AM
davidmccormick
A very annoying Inequality
Can anyone help me with this question:

Show that if f is continuous on the closed interval [0,1] then,

[IMG]file:///C:/DOCUME%7E1/OWNER%7E1.UNC/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/OWNER%7E1.UNC/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG]
http://reference.wolfram.com/chars/Integral.gif |f| < sqrt (http://reference.wolfram.com/chars/Integral.gif f^2)

the lower/upper limits of both integrals are 0 and 1 respectively.
• Nov 9th 2008, 06:30 PM
NonCommAlg
Quote:

Originally Posted by davidmccormick
Can anyone help me with this question:

Show that if f is continuous on the closed interval [0,1] then,

i guess this is the inequality you want to prove: $\int_0^1 |f(x)| \ dx \leq \sqrt{\int_0^1|f(x)|^2 \ dx}.$

this is a special case of Holder's inequality. however, it can be easily proved as an inequality on its own: clearly for any real number $\lambda$ we have: $\int_0^1(\lambda - |f(x)|)^2 \ dx \geq 0,$ which gives us:

$p(\lambda)=\lambda^2 - \left(2\int_0^1 |f(x)| \ dx \right) \lambda + \int_0^1 |f(x)|^2 \ dx \geq 0, \ \ \forall \lambda \in \mathbb{R}.$ thus the discriminant of the quadratic function $p(\lambda)$ must be $\leq 0,$ that is: $\left(\int_0^1 |f(x)| \ dx \right)^2 \leq \int_0^1 |f(x)|^2 \ dx. \ \ \Box$
• Nov 9th 2008, 06:35 PM
Mathstud28
Quote:

Originally Posted by NonCommAlg
this is a special case of Holder's inequality. however, it can be easily proved as an inequality on its own: clearly for any real number $\lambda$ we have: $\int_0^1(\lambda - |f(x)|)^2 \ dx \geq 0,$ which gives us:

$p(\lambda)=\lambda^2 - \left(2\int_0^1 |f(x)| \ dx \right) \lambda + \int_0^1 |f(x)|^2 \ dx \geq 0, \ \ \forall \lambda \in \mathbb{R}.$ thus the discriminant of the quadratic function $p(\lambda)$ must be $\leq 0,$ that is: $\left(\int_0^1 |f(x)| \ dx \right)^2 \leq \int_0^1 |f(x)|^2 \ dx. \ \ \Box$

I'm more impressed that you gleaned what the question was than your work.