Orthogonal functions with respect to a weight

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muzihc
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Homework Statement


Say functions f and g continuous on [a,b] and happen to be orthogonal with respect to the weight function 1. Show that f or g has to vanish within (a,b).

Homework Equations


f and g are orthogonal w.r.t. a weight function w(x) if
the integral from [a,b] of f(x)g(x)w(x)dx = 0.
(in this case w(x) = 1)

The Attempt at a Solution


I'm not sure what vanish means.
E.g. f(x) = x, g(x) = x^2 on some symmetric interval, say [-1,1] satisfies this, but I'm not sure how either function vanishes.

A more extreme example would be f(x) = cos(2x), g(x) = sin(2x), on say [-Pi,Pi]. How does either function
vanish? I'm not sure what that means, so I can't even begin a proof.
 
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muzihc said:

Homework Statement


Say functions f and g continuous on [a,b] and happen to be orthogonal with respect to the weight function 1. Show that f or g has to vanish within (a,b).

Homework Equations


f and g are orthogonal w.r.t. a weight function w(x) if
the integral from [a,b] of f(x)g(x)w(x)dx = 0.
(in this case w(x) = 1)

The Attempt at a Solution


I'm not sure what vanish means.
E.g. f(x) = x, g(x) = x^2 on some symmetric interval, say [-1,1] satisfies this, but I'm not sure how either function vanishes.

A more extreme example would be f(x) = cos(2x), g(x) = sin(2x), on say [-Pi,Pi]. How does either function
vanish? I'm not sure what that means, so I can't even begin a proof.

I think they just mean there is a value of x in (a,b) such that either f(x)=0 or g(x)=0. They certainly can't mean vanish identically.
 
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