A very annoying Inequality

**davidmccormick** · November 9th 2008, 10:58 AM

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]

|f| < sqrt (

f^2)

the lower/upper limits of both integrals are 0 and 1 respectively.

**NonCommAlg** · November 9th 2008, 06:30 PM

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: $\left(\int_0^1 |f(x)| \ dx \right)^2 \leq \int_0^1 |f(x)|^2 \ dx. \ \ \Box$

**Mathstud28** · November 9th 2008, 06:35 PM

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.

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