Regarding treating functions as 'vectors': if a function is not square-integrable, how would we define its norm & inner product, or would we even consider it a vector at all?

keenan

Good question. As mentioned in class, there are typically many different norms you can put on functions. For instance, beyond the L^2 norm one can define the so called L^p norm for any p >= 1:

||f||_p := ( \int_0^1 f(x)^p dx )^{1/p},

i.e., take the pth power of the function, integrate it, then take the pth root. Some functions that do not have a well-defined L^2 norm (because the integral diverges) may still have a well-defined L^p norm. Consider for instance the function

f(x) = 1/sqrt(x).

Are there values of p for which the L^p norm of this function is defined? Undefined?

yuxiangz

I see. So in this case p=2 would not work, but p=1 would.

Regarding treating functions as 'vectors': if a function is not square-integrable, how would we define its norm & inner product, or would we even consider it a vector at all?

Good question. As mentioned in class, there are typically

manydifferent norms you can put on functions. For instance, beyond the L^2 norm one can define the so called L^p norm for any p >= 1:||f||_p := ( \int_0^1 f(x)^p dx )^{1/p},

i.e., take the pth power of the function, integrate it, then take the pth root. Some functions that do not have a well-defined L^2 norm (because the integral diverges) may still have a well-defined L^p norm. Consider for instance the function

f(x) = 1/sqrt(x).

Are there values of p for which the L^p norm of this function is defined? Undefined?

I see. So in this case p=2 would not work, but p=1 would.

Right, exactly.