There is a theorem of Florian Vasilescu (not Florian Horia Vasilescu) saying that there is a sequence of continuous functions doing it. See Marcus, Din gindirea matematica romaneasca. Vasilescu proved it with multifunctions, but in early 70s Ionel Tevy proved it with Baire functions (and proved several related results). I will look up his papers and give you more details.

This might be it, Gabriel:

I. Tevy, Sur un theoreme de Florin Vasilescu. (Romanian) Zbl 0189.33802

Stud. Cercet. Mat. 21, 1173-1180 (1969),

but wish me luck in getting a copy...

I have his papers, he gave me hard copies last year when I got in touch with him, I just did not have them at work when I wrote earlier.

Last year I was thinking about something like this: Let A be the set of points of continuity for f. If A is finite then we can do it easy. If A is infinite then there is B = (x_n) a countable dense subset of A. Let P_n be the interpolating polynomial on x_1, x_2, ... , x_n. P_n converges to f on B. How about on A?

Along your lines, Gabriel: If the continuity set A is closed in [a, b], by Tietze’s theorem we can construct a continuous function g : [a, b] → R such that g(x) = f(x) for x ∈ A. And the sequence of polynomials given by Weierstrass' theorem for g will do the job for the function f and the set A (all continuity points of f).

This is Tevy's proof.

Since f is continuous on A included in [a, b], there is a Borel function g on [a, b] which coincides with f on A (he is quoting a theorem of extensions of Borel functions from vol 1 of Kuratowski from 1958). g being a Borel function is a pointwise limit of continuous functions. His result is more general, if f is class alpha Borel on A then it is the pointwise limit of a sequence of functions of class alpha Borel on [a, b]. Continuity is class 0.

There is a sequence of polynomial functions that converges to a continuous function f at every point where f is continuous. This is referred to as the Weierstrass approximation theorem. The theorem states that for any continuous function f : [a, b] → R and any ε > 0, there is a polynomial function p(x) such that |f(x) - p(x)| < ε for all x ∈ [a, b]. In simpler terms, the sequence of polynomial functions {p_n} converges uniformly to f on [a, b]. Thus, we can conclude that for any given function f : [a, b] → R, there is a sequence of polynomial functions that converges to f at every point where f is continuous.

Sure, Hamid, if the continuity set of our function is a compact interval. The latter is not necessarily true for all functions, e.g., the Thomae (Riemann) function. In fact, for any G_delta subset A on the real axis (i.e., a countable intersection of open sets), one can find a function whose continuity set is A.

The attached paper explains that our question has an affirmative answer for "f bounded on [a, b]" , e.g., for Riemann integrable functions on [a, b], or for functions having both one-sided limits (like monotone functions). Without adding further hypotheses of f, this is the best one can do.

George Stoica Hamid Reza Yazdani Gabriel T. Prajitura

Examples are included.

Yes, for any given function \(f: [a, b] \rightarrow \mathbb{R}\) that is continuous on the interval \([a, b]\), there exists a sequence of polynomial functions that converges to \(f\) at each point where \(f\) is continuous. This result is known as the Weierstrass approximation theorem.

Formally, the Weierstrass approximation theorem states that for any continuous function \(f: [a, b] \rightarrow \mathbb{R}\) and for any \(\epsilon > 0\), there exists a polynomial function \(P(x)\) such that \(|f(x) - P(x)| < \epsilon\) for all \(x\) in \([a, b]\).

In other words, given any continuous function, we can find a sequence of polynomial functions that get arbitrarily close to the function on the interval \([a, b]\).

An example of such a sequence of polynomial functions can be constructed using the Bernstein polynomials. The Bernstein polynomials are defined on the interval \([0, 1]\) and can be scaled and shifted to fit the interval \([a, b]\).

The \(n\)th Bernstein polynomial \(B_n(x)\) of a function \(f\) is defined as:

\[B_n(x) = \sum_{k=0}^n f\left(\frac{k}{n}\right) \binom{n}{k} x^k (1-x)^{n-k}\]

For any fixed \(n\), the Bernstein polynomial \(B_n(x)\) is a polynomial of degree \(n\) that approximates the function \(f\) on the interval \([0, 1]\). By scaling and shifting the interval, we can construct a sequence of polynomial functions \(P_n(x)\) that converges to \(f\) on the interval \([a, b]\).

Here's an example:

Let's say we want to approximate the function \(f(x) = \sin(x)\) on the interval \([0, \pi]\). We can construct a sequence of polynomial functions using the Bernstein polynomials:

\[P_n(x) = \frac{\pi}{2}B_n\left(\frac{2x}{\pi}\right)\]

As \(n\) approaches infinity, the sequence \(P_n(x)\) converges uniformly to \(f(x) = \sin(x)\) on the interval \([0, \pi]\). This means that for any \(\epsilon > 0\), there exists an \(N\) such that for all \(n > N\) and for all \(x\) in \([0, \pi]\), we have \(|f(x) - P_n(x)| < \epsilon\).

Therefore, the sequence of polynomial functions \(P_n(x)\) converges to \(f(x) = \sin(x)\) at each point where \(f\) is continuous on the interval \([0, \pi]\).

Hope it helps: partial credit AI

You did not understand the question. The function was NOT continuous on [a, b]. Not only that we know the theorem you talk about but we have been teaching it for 30 years. And of course it does not apply here.

AI is, as usual, useless.