Integral Inequality: Solving with Cauchy-Schwarz

[mohbb]

Homework Statement

Hi,

I must show that ($$\int cos(x)f(x)dx$$)^2 <= 2 $$\int cos(x)f(x)^2dx$$
(the integrals are from -pi/2 to pi/2)

The Attempt at a Solution

I know that I should use cauchey-schwarz inequality to solve this where <f,g> = $$\int f(x)g(x)dx$$ In this case i just set g(x) = cos x
Therefore i get
($$\int cos(x)f(x)dx$$)^2 <= $$\int f(x)^2dx$$ $$\int cos^2(x)dx$$ I calculated then integral of cos^2(x) which is 1/2(x + sin(2x)/2) since cos^2(x) = (1 + cos(2x))/2

However this leaves me with pi/2 to get:
($$\int cos(x)f(x)dx$$)^2 <= pi/2 $$\int f(x)^2dx$$

Howcome I am not getting the same answer as the one I should be proving, why does the question have a cos in the integral?

Thank you

 

[Dick]

Use a different inner product. Define <f,g> to be the integral of f(x)*g(x)*cos(x)*dx.