1. ## Gram-Schmidt method

do you integrate between 1 and 0 all the time.

Or is it what c is the space of the continous function. As it is -1 and 1?

do you integrate between 1 and 0 all the time.

Or is it what c is the space of the continous function. As it is -1 and 1?
The "Gram-Schmidt orthogonalization process" is a method of changing any basis in a vector space (with inner product) to a an orthonormal basis- a basis such that the norm of each vector is 1 and two distinct vectors have inner product 0.

Whether you integrate "between 1 and 0" or from -1 to 1 or whether you integrate at all depends upon what vector space you are talking about.

If you are talking about a set of continuous functions, you had better have some other way of defining norms and inner products because most continuous functions are not integrable at all!
What a remarkably silly thing to say! Of course all continuous functions (on a compact interval) are integrable. I must have been thinking of "differentiable".

It is common to define the "supremum norm" on continuous functions. If our vector space is the space of all functions continuous on [a, b] then we know that sup (f) (the least upper bound on |f(x)| for x in [a, b]) exists for every such f. But there is no "inner product" that gives that norm so "Gram- Schmidt" cannot be used for that set.

If you have the set of functions such that $\int|f| dx$ is defined on [a, b], then we could define the norm as $\int |f(x)|dx$ (Often called " $L^1[a,b]$") but, again, there is no "inner product" associated with that norm so "Gram-Schmidt" cannot be used.

The set of "square integrable functions" on a set, $L^2[a, b]$, such that $\int_a^b f^2(x) dx$ exists, also has the property that $\int_a^b f(x)g(x) dx$ so that we can define the "inner product" of f and g in that way. That gives both an inner product and norm so that Gram-Schmidt can be used.

But the problem is that "Gram-Schmidt" is applicable in much more examples than you seem to understand- so it is hard to know exactly what you are asking or what sort of response you would understand.

3. Originally Posted by HallsofIvy
The "Gram-Schmidt orthogonalization process" is a method of changing any basis in a vector space (with inner product) to a an orthonormal basis- a basis such that the norm of each vector is 1 and two distinct vectors have inner product 0.

Whether you integrate "between 1 and 0" or from -1 to 1 or whether you integrate at all depends upon what vector space you are talking about.

If you are talking about a set of continuous functions, you had better have some other way of defining norms and inner products because most continuous functions are not integrable at all!

It is common to define the "supremum norm" on continuous functions. If our vector space is the space of all functions continuous on [a, b] then we know that sup (f) (the least upper bound on |f(x)| for x in [a, b]) exists for every such f. But there is no "inner product" that gives that norm so "Gram- Schmidt" cannot be used for that set.

If you have the set of functions such that $\int|f| dx$ is defined on [a, b], then we could define the norm as $\int |f(x)|dx$ (Often called " $L^1[a,b]$") but, again, there is no "inner product" associated with that norm so "Gram-Schmidt" cannot be used.

The set of "square integrable functions" on a set, $L^2[a, b]$, such that $\int_a^b f^2(x) dx$ exists, also has the property that $\int_a^b f(x)g(x) dx$ so that we can define the "inner product" of f and g in that way. That gives both an inner product and norm so that Gram-Schmidt can be used.

But the problem is that "Gram-Schmidt" is applicable in much more examples than you seem to understand- so it is hard to know exactly what you are asking or what sort of response you would understand.
Thanks for writing that out for me.

I have to let C[-1, 1] to be the space of a continuous function f: [-1,1} -> C with the inner product <f,g> = the integratin of f(x)g(x) dx between -1 and 1 and g(x) is the complex conjugate.