# L2 norm = +Infinity

1. Jun 3, 2009

### mnb96

Hello,
I have a (infinite dimensional) vector space and defined an inner product on it.
The vectors element are infinite sequence of real numbers $$(x_1, x_2,\ldots)$$.
The inner product has the common form: $$x_iy_i$$

The problem now is that the vectors have an infinite number of elements, so the L2-norm of many vectors would be eventually equal to +Infinity.

- Is that admitted?
- How can one define an orthonormal base for such a space?

2. Jun 3, 2009

### sparkster

Are the vectors sequences that are eventually zero?

3. Jun 3, 2009

### DrGreg

You mean $$\sum^{\infty}_{i=1}x_iy_i$$
Yes, provided you restrict yourself to the subspace of sequences that have a finite norm, that is, for which the infinite series $\sum x_i^2$ converges to a finite limit. Ignore the series that do not converge -- those sequences are not elements of the inner product space.

How would you answer this for the space of sequences of length N? It's almost the same answer here.

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

4. Jun 4, 2009

### mnb96

I am not sure about what I am about to say, but as far as I understood, the inner-product that goes to +Infinity is always admitted: there is nothing in the definition of inner-product that prevents it to be so.

Since an inner-product-space is apparently just a vector space with an inner product, we have to admit that also those series $$\sum^{\infty}_{i=1}x_iy_i$$ 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 Hilbert-Space (complete metric).

Now, if we have an Hilbert-space, 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. Jun 4, 2009

### HallsofIvy

Staff Emeritus
I think there is confusion here as to what, exactly L2 means.

The space l2 (small l) is defined as "the set of all infinite sequences {an} such that $\sum a_n^2$ is finite". The l2 norm is then defined as $\sum a_n^2$ which is now guaranteed to be finite. And that norm can be derived from an "inner product" $\{a_n\}\cdot\{b_n\}= \sum a_nb_n$ which can be shown to always exist.

The space L2 Is defined as the set of functions, f(x), is defined to be the set of functions, defined on some set A, such that $\int_A (f(x))^2 dx$ is finite. Now the "L2" is defined to be that integral which is not guaranteed to be finite. And, again, it can be derived from the inner product, $f\cdot g= \int_A f(x)g(x) dx$.

That is, the "L2 norm" and "l2" are not defined independently of the set of "vectors".

6. Jun 4, 2009

### mnb96

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 $$L_{2}$$ 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?

Last edited: Jun 4, 2009
7. Jun 4, 2009

### maze

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.

Last edited: Jun 4, 2009