Please see attached my take on this problem.

Really, its answer is established if f belong to L^1[a, b].

Maybe you think about "continuous-like" functions? It looks like the Bernstein approximation result. Any regulated function (i.e., having one-sided limits at each point) is a uniform limit of step-functions. They are limits of polynomials you asked... (see https://www.emis.de/journals/DM/v16-1/art14.pdf )

Weierstrass Approximation Theorem

1. the function f(x) is a continuous real-valued function

2. on the interval [a,b]

3. if epsilon > 0

Then, there exists a polynomial p on the interval such that

|f(x) - p(x)| < epsilon

It does look like Bernstein's theorem, with two differences: f is not required to be bounded by hypothesis, and the conclusion is not asking for uniform convergence.

Hmmmmm... Sir, you have made a point. With two differences? Sir, permit me to know what you mean by this?

Dear Isaac,

It means that I will be satisfied with pointwise convergence of the polynomial functions converging to f at each point where f is continuous. (I would not look/hope for uniform convergence on the continuity set of f).

You think about continuous convergence?

No, just plain pointwise convergence:

https://en.wikipedia.org/wiki/Pointwise_convergence

I am not sure how continuity relates to the question. It may, but there are other factors.

Take any function and any countable subset of the domain. You can construct a sequence of interpolating polynomials that will converge pointwise to the function at each of these points. It was irrelevant if the function was continuous or not at those points. I guess here the relevant fact is the size of the set (countable).

Take the function = 0 on the irrationals and 1 on the rationals (on [0, 1]). The sequence (1/n) x converges uniformly to it on the irrationals. Here the relevant fact was that there is an everywhere continuous function equal to this one on the part of the domain where we approximate.

One more thing: you can relax the question to ask if there are continuous functions that approximate pointwise. If you can do it with continuous functions you can also do it with polynomials. Not a big deal, but it may just give more freedom to construct functions.

Dear George

The following paper is shown your version of the Weierstrass approximation theorem that it is not right in generality.

"Faber, G., Über die interpolatorische Darstellung stetiger Funktionen, Jahresber. Deutsch. Math. Verein., 23 (1914), 190–210."

Thanks

Thank you Mehdi, I translated that paper using my rusty German language knowledge, but the findings therein refer more to uniform interpolation of continuous function, which I am not looking for in our exercise. Good paper, though...

Sorry George, in really, it is rare which the interpolating polynomials converge to continuous functions.

Please, you could also see the following paper

"Trefethen, Lloyd N., Six myths of polynomial interpolation and quadrature, Mathematics Today, 47 (4) (2012), 184–188"

Good paper Mehdi, about "myths". We'll keep on trying, who knows...

This paper is about idle thinking of interpolating polynomials.

I hope it is benefit for your aim.

Dear Gumrah,

This is why I formulated my request as a question.

Do you have any counterexamples to my question?

Thank you for your interest!

George

Dear Gumrah,

I have to stay on a compact interval, otherwise there might be some complications. See, e.g., Rogier Brussee's answer (2nd from above) in:

https://www.researchgate.net/post/Can_be_replaced_closed_interval_a_b_by_open_interval_a_b_in_Weierstrass_Approximation_Theorem2

Sincerely,

George

Dear Gumrah, please see attached.

Dear George,

thank you for sharing a nice question. I like it since I could find a reduction of the question:-)

This is the main step for reduction (AFTER A CORRECTION)

>> For any given function f : [a, b] → R, there exists a sequence of CONTINUOUS (instead of POLYNOMIAL) functions converging to f at each point where f is continuous. (Note that we did not ask the convergence to be uniform).

Very elegant arguments, dear Joachim! - Thank you. I will keep this question open because, as you said, measurability is hard to get rid off. On the other hand, I have my eyes on Tietze's extension theorem on closed subsets. The problem is, of course, to suitably guess that particular closed set.

Dear Domsta

Even it is not true for any measurable function!!!

You see a real analysis book, such as ruiden, aliprantic, and so on.

Thanks

Really, its answer is established if f belong to L^1[a, b].

Please see attached my take on this problem.

Dear George Stoica ,

I read your proposed solution, in the w.pdf document. I do not see why the limit defining f1 on the boundary of Cf would exist. For example f(x)=1/x with f(0)=0 does not satisfy this.

True, Omran, as usual I've been too optimistic. I'll try better, maybe adding from the start the hypothesis that f is bounded on [a, b]. Be right back.

Sorry, George, but the boundedness will not help in your construction of f1

Example: the unit Heaviside step function: how to define the value at the origin by the limit of values at points in the nbhd to get a continuous function on R?

Example 19. “A function with a dense set of points of continuity, and a dense set of points of discontinuity no one of which is removable.”

Example 22. “A function whose points of discontinuity form an arbitrary given closed set.”

Example 23. “A function whose points of discontinuity form an arbitrary given F_σ set.”

Example 24. “A function that is not the limit of any sequence of continuous functions.”

(See Examples 19, 22, 23 and 24, Chapter 2, pages 28, 30 and 31 in the book: B.R. Gelbaum and J.M.H. Olmsted, Counterexamples in Analysis, Dover Publications, Inc. Mineola New York, 1964.)

“Note 2. According to a theorem of Blumberg, any given function f : [a, b] → R is continuous on a dense subset of [a, b]. The above problem goes a step further and says that for any given function f : [a, b] → R, there exists a sequence of polynomial functions converging to f on a dense subset of [a, b].” (Posted by George Stoica in attached file, 5 days ago).

Let A = {x1, x2, …, xn,…} be any countable (possible finite) subset of [a, b], and let f : [a, b] → R be a function such that f(xk) = yk for all k = 1, 2,…. For each fixed n ϵ {1, 2,…}, let Ln(x) be the interpolation polynomial in the Lagrange form of degree n – 1 such that Ln(xk) = f(xk) = yk for all k = 1, 2,…, n. Then for any fixed k ϵ {1, 2,…} we have Ln(xk) = f(xk) = yk for all k = n, n + 1, n + 2,…, and hence, Ln(xk) → f(xk) as n → ∞, i.e., the sequence of polynomials Ln(x) converges to f on a set A.

Consequently, if the set Cf of continuity points of a function f is countable, then the answer to the proposed question is positive.

Further, Examples 2 and 14, resp. on pages 22 and 27 in Chapter 2 of the book: B.R. Gelbaum and J.M.H. Olmsted, Counterexamples in Analysis, Dover Publications, Inc. Mineola New York, 1964.) are given as follows.

Example 2. A function continuous at one point only.

Let f: R: → R be a real function defined as: if f(x) = x if x is a rational number, and f(x) = - x if x is a irrational number. Obviously, the function f is continuous only at a point x = 0. Clearly, every sequence (Pn) of polynomials whose free coefficients are zero satisfies the required condition.

Example 15. A function continuous at every irrational point and discontinuous at every rational point.

Let f: R: → R be a real function defined as follows. Let f(0) = 1. If x = m/n is a nonzero rational number written in reduced form, i.e., n > 0 and m and n are relatively prime integers, then define f(m/n) = 1/n. Otherwise, if x if is a irrational number, let f(x) = 0. Then it follows easily that the function f is continuous at every irrational point and discontinuous at every rational point. Clearly, every sequence (Pn) of polynomials such Pn(x) ≡ 0 for each n > n_0 with some n0 satisfies the required condition.

Question. Do there exist some other (nontrivial) sequences (Pn) of polynomials such that for every irrational number x, Pn(x) → 0 as n → ∞?

Remark. Notice that the answer is positive if the set of irrational numbers is replaced by any finite or countable infinite set of real numbers, assume a set A = {x1, x2,…, xn,…}. Then define the sequenc (Pn) of polynomials as follows.

P1(x) = x – x1, P_2(x) = (x – x1)(x – x2),…, Pn(x) = (x – x1)(x – x2)...(x – xn),….

Since for any fixed k, Pn(xk) = 0 for all n ≥ k, we have Pn(xk) → 0 as n → ∞.

Dear Romeo Meštrović ,

your contribution with the sofisticated examples is very interesting and of high educational usefulness. Dziękuję [PL]

PS: How would "dziękuję" be in your native language? Is this црногорски jeзик? Is the name of your country abbreviated on cars as MNE ?