# Interchaning Limits and Inner Products

1. Oct 23, 2013

### logarithmic

Everything here is in a Hilbert space. If $x_n\to x$ and $y_n\to y$ in norm, then under what conditions does
$<x_n,y_n>\to <x,y>$?

Is this always true, and why?

Does anyone have a source?

2. Oct 23, 2013

### CompuChip

I think the answer is "yes, always"; for the following reason.
The inner product is continuous function in both arguments, and for continuous functions you can "pull" the limit inside the function (i.e. limn→∞ f(xn) = f(limn→∞ xn)).

Does this give you enough of a handle for a more rigorous approach?

3. Oct 23, 2013

### economicsnerd

What you're asking is exactly whether $\langle\cdot|\cdot\rangle:\mathcal H \times \mathcal H \to \mathbb C$ (or $\mathbb R$) is continuous. The answer is yes. To start you off on seeing why, notice that $$\langle x_n,y_n\rangle-\langle x,y\rangle = \langle x_n,y_n\rangle- \langle x_n,y\rangle + \langle x_n,y\rangle- \langle x,y\rangle$$ and go from there.

4. Oct 23, 2013

### R136a1

You will need the Cauchy-Bunyakovski-Schwarz inquality.

5. Oct 26, 2013

### epr1990

Exchanging limits and anything else, i.e. derivatives, sums, integrals depends on whether or not a sequence of functions is uniformly convergent. Since in Hilbert space, the inner product is either a sum for discrete or an integral for continuous cases, such a result is dependent on whether or not the sequences of functions x_n and y_n converge uniformly to x and y respectively. That is the answer to your question.

More explicitly, if you have two sequence of functions, x_n and y_n which are continuous on some set and converge uniformly to x and y, then x and y will also be continuous on those sets. Likewise, if you take the inner product of two sequences of functions take the limit, you could interchange the limit and the integral (or sum) if the integral (or sum) converges for each n and if the sequences of functions are uniformly convergent.

Note: the latter result doesn't even need the functions to be continuous, and also this has nothing to do with whether or not the inner product is a "continuous function"... I actually don't think that is a true statement in general. It is however sesquilinear (linear in its first argument and antilinear in its second), which is basically consequence of the fact that its basically sum and conjugate symmetry. If the codomain is the set of real numbers (< . , . >: HxH --> R), then conjugate symmetry becomes symmetry, and sequilinearity becomes bilinearity. Regardless, because of this linearity, the basic rules of limits apply which are exactly what I just stated.

As for a source, proofs of these statements can be usually found in any elementary book on real analysis. Many will tell you to check out Rudin, or someone like that, I personally am very fond of Real Mathematical Analysis by Charles Chapman Pugh. However, I don't know if you could find a statement of this exact result. Most real analysis books will begin with real analysis of functions of one variable and build up to how and when to interchange things like this, and then generalize to very abstract spaces and expect you to be able to infer and apply results like this to those more abstract spaces. However, generalizations of the basic concepts to other types of metric spaces be easily arrived at once the basic idea is understood.

Last edited: Oct 26, 2013
6. Oct 26, 2013

### Robert1986

I'm not sure this is correct. For example, functions can converge in L2-norm with out converging uniformly. Indeed, functions can converge in L2-norm with out even converging ae.

True, but so what? This is not relevant in this case.

This is true, but you don't need hypotheses this strong.

What?!?!? This is exactly what it means for the inner product to be continuous!! Ie, since if f_n converges to f in norm (where the norm is induced by the inner product), then the inner product of f_n against all g converges to <f,g>. This is equivalent to being continuous in the norm topology.

It is. And it can be seen as being true from the second reply in this thread.

7. Oct 26, 2013

### R136a1

Right. Uniform convergence has nothing do with this exercise here. Interchanging limit and integral can even be done without uniform convergence, for example see the monotone or dominated convergence theorem.

8. Oct 26, 2013

### pwsnafu

With respect to the derivative, that is misleading. The sequence of functions $f_n$ can converge pointwise. It's the derivatives $f'_n$ that needs to converge uniformly.

Also only the Riemann integral cares about uniform convergence, but Riemann doesn't form Hilbert spaces.

9. Oct 26, 2013

### Robert1986

I don't think this is quite true. For example, Egorov's Theorem is a theorem about uniform convergence. Most of the time, though, this is used on a compact set, and as long as the functions are also riemann integrable, we can just use riemann integrals, but this isn't always the case and can be fairly limiting at times.

10. Oct 27, 2013

### pwsnafu

Almost uniform convergence, technically.

But does that form a Hilbert space?

Remember Lebesgue theory is on the relative complement of set with zero measure, while Egorov is limited to the relative complement with positive measure.

Edit: maybe that space is dense in L2? Can you verify?

Last edited: Oct 27, 2013
11. Oct 27, 2013

### Robert1986

Well, I was just pointing out that uniform convergence is still important when it comes to lebesgue integral, even though you are certainly correct that uniform convergence doesn't play nearly the same role in lebesgue integration as in riemann.

12. Oct 27, 2013

### pwsnafu

Oh okay, gotcha. Yeah I agree with you in that case.