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(a) Give an example of an orthonormal system on [0,1] consisting of nonnegative continuous functions. (b)Prove that there is no orthonormal system on [0,1] consisting of continuous positive functions.

I have no idea how to start so any solution or hint will be much appreciated! Thank you in advance!

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    what is an orthonormal system? – 2017-02-22
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    Perhaps you are looking for an infinite set of nonnegative functions on [0,1] that are orthonormal with respect to the inner product $(f,g) = \int_0^1 f(x)g(x) dx$? – 2017-02-22
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    What is the inner product you are talking about ($L^2$ maybe)? I think you can use the Gram–Schmidt process ([this](https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process)). Unfortunately, i don't have a pen and paper with me so I cannot yet help you...... – 2017-02-22
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    Yes @Merkh. We need the integral from 0 to 1 to be zero when you use different functions and f and g (this is orthogonal). If it is orthonormal the integral from 0 to 1 of f^2 must be 1. – 2017-02-22

4 Answers 4

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Part $b$ is easy because if $f$ and $g$ are positive functions then $fg$ is a continuous positive function and we must have $\int\limits_{0}^{1}fg(x)dx> 0$

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    @JuliánAguirre: The functions' support is $[0,1]$. They are positive over their support... At least I try to interpret the answer above like that. – 2017-02-22
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The second question has been answered.

As far as the first one:

Consider isosceles triangles over $[0,1]$. Let the height of the first triangle over $(\frac12,1]$ be 4. Let the height of the second triangle over $[\frac14,\frac12]$ be $8$ and so on... The graph of the functions be given by the listed triangles and $0$ outside of their base.

The integral of these functions is one. The product of any two of them is zero. They are continuous...

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Part a) Define $f_1$ as a $C^\infty$ normalized function with its support in $[0,1/2].$ Define $f_2$ as a $C^\infty$ normalized function with its support in $[1/2,3/4].$ Keep going with this.

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    These functions *do* exist, by the way. Its a common exercise in the theory of distributions to show that compactly supported $C^\infty$ functions exist. I'd suggest trying it out yourself. – 2017-02-22
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You could build a set of polynomials that would work.

$f_1(x) = 1\\ f_2(x) = a (x-\frac 12)$

Centering your polynomials in the middle of the interval will make things easier in the long run.

now calibrate a such that:

$\int_0^1 f_1(x) f_2(x) \ dx = 0$ and $\int_0^1 f_2^2(x) \ dx = 1$

$f_2(x) = 12(x - \frac 12)$

etc.

But, this sounds like work

The trigonometric functions are orthogonal when evaluated over a full period, and make for a very nice basis.

$a_0 = 1, a_n = 2 cos 2n\pi, b_n = 2 \sin 2n\pi$

as for b) show that

$\int_0^1 f_1(x) f_2(x) \ dx \ne 0$ if both $f_1(x)$ and $f_2(x) > 0$ for all $x$ in the interval.

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    But your f2 is not always nonnegative – 2017-02-23
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    No, it's not. $f_2(0) = -6$ – 2017-02-23