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L2 norm = +Infinity 
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#1
Jun309, 09:20 AM

P: 626

Hello,
I have a (infinite dimensional) vector space and defined an inner product on it. The vectors element are infinite sequence of real numbers [tex](x_1, x_2,\ldots)[/tex]. The inner product has the common form: [tex]x_iy_i[/tex] The problem now is that the vectors have an infinite number of elements, so the L2norm of many vectors would be eventually equal to +Infinity.  Is that admitted?  How can one define an orthonormal base for such a space? 


#2
Jun309, 02:39 PM

P: 153

Are the vectors sequences that are eventually zero?



#3
Jun309, 04:00 PM

Sci Advisor
PF Gold
P: 1,850

(By the way, this post should really be in the Linear & Abstract Algebra forum!) 


#4
Jun409, 03:01 AM

P: 626

L2 norm = +Infinity
I am not sure about what I am about to say, but as far as I understood, the innerproduct that goes to +Infinity is always admitted: there is nothing in the definition of innerproduct that prevents it to be so.
Since an innerproductspace is apparently just a vector space with an inner product, we have to admit that also those series [tex]\sum^{\infty}_{i=1}x_iy_i[/tex] which do not converge are allowed. So, the vectors with Infinite norm are still in the inner product space. If you add the requirement that those series have always to converge, then you are defining an HilbertSpace (complete metric). Now, if we have an Hilbertspace, it is probably easier to define an orthonormal basis. I am not sure it is possible to define always an orthonormal base for inner products, as I can't see how you could (for example) normalize the squared integral of a sinusoid extendind through the whole real line. 


#5
Jun409, 05:46 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

I think there is confusion here as to what, exactly L_{2} means.
The space l_{2} (small l) is defined as "the set of all infinite sequences {a_{n}} such that [itex]\sum a_n^2[/itex] is finite". The l_{2} norm is then defined as [itex]\sum a_n^2[/itex] which is now guaranteed to be finite. And that norm can be derived from an "inner product" [itex]\{a_n\}\cdot\{b_n\}= \sum a_nb_n[/itex] which can be shown to always exist. The space L_{2} Is defined as the set of functions, f(x), is defined to be the set of functions, defined on some set A, such that [itex]\int_A (f(x))^2 dx[/itex] is finite. Now the "L_{2}" is defined to be that integral which is not guaranteed to be finite. And, again, it can be derived from the inner product, [itex]f\cdot g= \int_A f(x)g(x) dx[/itex]. That is, the "L_{2} norm" and "l_{2}" are not defined independently of the set of "vectors". 


#6
Jun409, 05:56 AM

P: 626

thanks for the clarification!
Now, in this context, could you please explain what happens if we consider the space of the real functions obtainable by sum of complex sinusoids? The complex sinusoids (from Inf to +Inf) have infinite norm. This means we are not in [tex]L_{2}[/tex] anymore. However the sinusoids are still orthogonal(?), so we must conclude they are a basis for some space, but what space? In other words, when we simply take a Fourier Transform of a function from Inf to +Inf, what are we actually doing? 


#7
Jun409, 07:07 AM

P: 655

To ease explanation, consider the complex exponentials e^{ikx} instead of sines and cosines, and consider the complex exponential version of the fourier transform.
If you give it a bit of thought, you will realize that the fourier transform of e^{ikx} is not actually a function, but rather a delta "function" (really, the delta distribution). Why? e^{ikx} is a perfect wave of a single frequency, so it's Fourier transform has all of the weight concentrated at a single point k, and no weight at any other frequencies. Therefore the natural space to think about Fourier transforms of things like e^{ikx} is a space of distributions. The space commonly used is the dual space to the Schwarz space, and then the Fourier transform is F:S'>S' rather than F:L2>L2. The complex exponentials might form a Schauder basis for S', though I highly doubt it. They certainly don't form a Hamel basis. This is actually an interesting question. 


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