Polynomial Functions Are Dense in Continuous Function Space

2. Are the analytic functions dense in the space of continuous functions?

Are the analytic functions dense in the space of continuous functions?

1. As was mentionned in a previous answer before the question was put on hold, Stone-Weierstrass's theorem states that polynomial functions are dense in the space of continuous functions $f: [a,b] \to \mathbb R$ for the uniform convergence, which implies $L^2$ density. In fact, since continuous functions are dense in $L^2$, this means that any measurable function $f: [a,b] \to \mathbb R$ with $\int_{a}^b |f|^2 <+\infty$ can be $L^2$-approximated by polynomials, no matter how discontinuous $f$ is.

2. This still works for functions defined over an infinite interval, but more work is required (it is not an immediate consequence of Stone-Weierstrass).

3. If $U \subset \mathbb C$ is an open set, then the space of complex analytic functions $f: U \to \mathbb C$ that is also in $L^2(U)$ is closed for the $L^2$ norm (it is called the Bergman space $A^2(U)$). In particular, it is not dense in the space of continuous complex-valued functions defined on $U$.

4. I am not sure if this case is also of interest to you, but if $K \subset \mathbb C$ is a compact without interior, you can ask whether complex-valued continuous functions on $K$ can be uniformly approximated by restrictions to $K$ of analytic functions. This is a more delicate question, which depends on $K$. It is known for example that if $K$ disconnects the plane in at most finitely many components, the answer is yes; same thing if $K$ has zero Lebesgue measure.

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Comments

• 1. We can consider functions defined on the real or complex numbers in any finite number of dimensions. The functions are defined everywhere. They need not have limited support. They may be defined over open or closed subsets of R^n or C^n (typing on my phone so avoiding latex) or everywhere.

  2. In defining continuous functions, continuity is considered in the context of the topology induced by the Euclidean metric.

  3. Distance between functions can be taken to be the distance in the L2 norm (the integral over the entire domain of the squared difference between two functions).

  Which of these assumptions, if changed, might change the answer to the question? For example, we might want to be able to approximate continuous functions with analytic functions. But we might also wish to do this confidently for as wide a class of functions as possible: for instance, what about piecewise continuous functions with finite or countable points of discontinuity? We can consider widening the class of functions to include either of the latter two. Moreover, we might want to impose the strictest convergence criteria as possible for the approximations. For example, we might want to strengthen the requirement from closeness in L^2 to closeness that is uniform or close to uniform (e.g. uniform but for a finite set of points, i.e. "almost everywhere"), where closeness is viewed in the sense of convergence of a sequence of successively better approximations. So consider the stated (numbered) assumptions, but feel free to comment on whether the answer would still hold for more favorable assumptions given that the goal is to approximate functions as stated.

  • Before the question can be answered, you must specify which topology you consider on the space of continuous functions defined on $\Bbb R$

  • @SherifF. "Context" here (in my opinion) would be a couple of crucial technical details: 1. Are your functions defined everywhere, or only on a closed, bounded (i.e., compact) set? 2. How do you measure "closeness" of two functions? (The supremum of the magnitude of their difference? The integral of the square of their difference...?) (It does help to know that you're an engineer by training, but I agree with Hagen that as stated your question cannot be answered mathematically. We need the specifics of your mathematical hypotheses. :)

  • Sure, I understand. In most applied contexts, you consider continuity in the topology induced by the Euclidean metric. That is the natural context. Continuous function spaces can include functions defined everywhere (which of couse includes those defined on a compact subdomain of R^n or C^n). Functions are close to each other if they are close "almost everywhere" if that makes sense. Alternatively, we can consider their distance in L2 norm. Are these equivalent? Does the answer differ depending on these various assumptions? If so, how? That can be part of the answer.

  • The Euclidean metric on $\mathbb{R}^n$ or $\mathbb{C}^n$ does not directly impute a topology on the spaces of continuous or of analytic functions. Simply knowing how far apart points are does not say how far apart functions are. If you want to use Math.SE to its fullest potential, give a full problem statement in the body of the Question. Here you limited the problem formulation to what you could reasonably get into the title.

  • Above, I mentioned two (possibly equivalent) notions of distance between functions.

  • Perhaps you should start with asking whether the two notions of distance between functions are equivalent (they are not; the $L^\infty$ norm gives the maximum absolute difference between two functions, and is not finitely-defined for the spaces of continuous or analytic functions on an unbounded domain).

  • Let us consider L^2 not L^infinity. So it's the integral of the squared difference between functions.

  • I edited the question to define a notion of continuity, a notion of the domain over which functions are defined, and a notion of distance between functions, which induces a topology on the space of functions. Further, I specify that the functions are real or complex valued. If this is not sufficiently specific, please explain why. If it remains problematic, I will delete the question.

  • @SherifF. Analytic functions can be defined over open intervals (real-analytic functions) or over open subsets of $\mathbb{C}$ (complex-analytic functions). The situation also depends on whether the interval is finite or infinite. You should also include in your question what you mentionned in the comments, that you are asking about density for the $L^2$ norm (once again, things are different with different norms). If you do all that I will reopen the question.

  • Please see the latest edit to the question. Does it address all the concerns?

  • @Glougloubarbaki, any thoughts on removing that hold?

  • @SherifF. I voted to reopen

• Thank you. In your third situation, can you clarify what you mean by the statement that the space of these functions is closed for the $L^2$ norm? Closed over what operation? And do you mean to state that since any finite linear combination of them will also be $L^2(U)$ (i.e. closure under addition and scalar multiplication) implies that they can only approximate functions that are also $L^2(U)$, or did I misunderstand?

• @SherifF. I mean that any $L^2$ limit of a complex analytic function will be itself complex analytic. Therefore, any continuous function that is not analytic cannot be $L^2$ approximated by analytic functions on $U$

• Do you have a reference for point number 4 (specifically the measure zero case)? This is the one that I am most interested in.

• Never mind, I found the names of the references (Mergelyan's theorem and the hartogs-rosenthal theorem).

Recents

What is the matrix and directed graph corresponding to the relation $\{(1, 1), (2, 2), (3, 3), (4, 4), (4, 3), (4, 1), (3, 2), (3, 1)\}$?

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