Does there exist such a function?

Question

Is there a function f: R --> R such that for every point x in R, lim(y -->x) f(y) exists but is not equal to f(x).

Answer 1

Benoit's proof is seriously flawed. It proves only that ∀ε>0 ∃z ∃δ: |y-z|<δ → |f(y)-f(z)| < ε. Which is not the same as proving ∃z: ∀ε>0 ∃δ: |y-z|<δ → |f(y)-f(z)| < ε, which is what he needs to prove to show [y→z]lim f(y) = f(z). Indeed, his summary of his proof, namely that "In short... if the limit is not f(x) for some x, then the limit has to be equal to f(z) for all z within a neighborhood of x" is demonstrably false -- consider the popcorn function. The limit at every point is zero, and it is unequal to the value of the function at every rational number, but there is no neighborhood of _any_ rational number in which the limit is equal to f(z) for all z in the neighborhood.

In fact, the popcorn function tells us that we cannot rely only on the properties of limits in proving this theorem, since the restriction of the popcorn function to Q actually does satisfy the property that for every point x in Q, [y→x]lim f(y) exists but is not equal to f(x). We must rely on one additional property, namely the fact that the real numbers are uncountable. First, a lemma:

Lemma 1: If ∀x∈[0, 1], [y→x]lim f(y) exists, then ∀ε>0, there are only finitely many points in [0, 1] such that |f(x) - [y→x]lim f(y)| > ε.

Proof: suppose the contrary -- that ∃ε>0 such that infinitely many points in [0, 1] satisfy |f(x) - [y→x]lim f(y)| > ε. Let P be the property that there are infinitely many such points in the given interval that satisfy |f(x) - [y→x]lim f(y)| > ε. Clearly, if an interval [a, b] satisfies P, then at least one of [a, (a+b)/2] and [(a+b)/2, b] satisfy P. Therefore, we can construct a sequence of nested subintervals as follows: let I_0 = [0, 1], and if I_n = [a, b], then I_(n+1) = [a, (a+b)/2] if it satisfies property P, otherwise [(a+b)/2, b]. These form a sequence of nested closed intervals, and so by the nested intervals theorem, there is a point c such that c∈∩I_n. Now, let L = [y→c]lim f(y), then ∃δ>0 such that 0<|x-c|<δ → |f(x)-L|<ε/2. Further, if 0<|x-c|<δ, then ∃d>0 (specifically, d=min(|x-c|, |x-c+δ|, |x-c-δ|)) such that 0<|y-x| ε and thus at least one such that x≠c. So there is some x satisfying |f(x) - [y→x]lim f(y)| > ε such that 0<|x-c|<δ and thus |f(x) - [y→x]lim f(y)| < ε, a contradiction! Therefore, ∀ε>0, there are only finitely many points in [0, 1] such that |f(x) - [y→x]lim f(y)| > ε.

Note that the contradiction is superficially similar to Benoit's proof, namely that of showing that for x sufficiently close to c [y→x]lim f(y) is near f(x), but we did not have to show that they were actually equal (which is good, because we know of counterexamples where they are in fact unequal).

Now, on to the main theorem, which practically falls out of the lemma:

If ∀x∈[0, 1], [y→x]lim f(y) exists, then there are at most countably many points on [0, 1] such that f(x)≠[y→x]lim f(y)

Proof: ∀n∈N, let A_n be the set of all points in [0, 1] such that |f(x) - [y→x]lim f(y)|>1/n. Let A=∪A_n. Clearly, A is the set of all points such that f(x)≠[y→x]lim f(y). For each n, A_n is a finite set per lemma 1, so A is a countable union of finite sets and therefore countable.

Now, there are uncountably many points on [0, 1], so if ∀x [y→x]lim f(y) exists, then you cannot have that ∀x, f(x)≠[y→x]lim f(y). And if you can't do this on [0, 1], you certainly can't do it on the real line. Therefore, no function conforming to your specifications exists.

Answer 2

No. Proof by contradiction... Let's assume there is one.

Given x E R, and e>0. Let L(x) = lim(y→x) f(y).
Since L(x) exists, there is a number d(e,x) > 0 such that
|f(y) - L(x)| < e for every y such that 0 < |y-x| < d(e,x).

Let z = x + d(e/2,x)/2. Then z falls within the required neighbourhood of x so that we have the inequality
|f(z) - L(x)| < e/2.

Now let y be such that 0 < |y-z| < d(e/2,x)/2, we have
0 < |y - x| < |y - z| + |z - x| (triangle inequality)
< d(e/2,x)/2 + d(e/2,x)/2 = d(e/2,x).
So y is also within the neighbourhood of x such that
|f(y) - L(x)| < e/2.

And so
|f(y) - f(z)| < |f(z) - L(x)| + |L(x) - f(y)| (triangle inequality)
< e/2 + e/2 = e

In other words, L(z) = lim(y→z)f(y) = f(z) which contradicts the hypothesis.

In short... if the limit is not f(x) sor some x, then the limit has to be equal to f(z) for all z within a neighbourhood of x.

Answer 3

While it is true that uniform continuity does imply continuity, the converse is not true. For example if f(x) is unbounded then f(x) is not everywhere uniformly continuous. Since the existence of uniform continuity requires (independently of x) that: For all ε > 0, there exists a δ such that for two points x1 and x2, such |x1-x2|< δ, we must have have | f(x1) − f(x2) | < ε. Let us consider f(x) = x^3 Choose an ε > 0. For any positive number δ, we have f(x + δ) − f(x) = 3x^2*δ + 3x*δ^2 +δ^3, and for x sufficiently large this exceeds ε. However, for any subset of R that is the domain of a bounded function f(x), we do have uniform continuity. For any continuous f(x) there must be such an interval, that is where f(x) is bounded. Therefore the answer to the problem is NO! No such function exists. .