How can we prove this sequence of functions converge uniformly?

Question

I'd like some help to prove this:

Let f_n be a sequence of monotonic functions defined on [a, b] that converges to a continuous function f. Then, the convergence is uniform.

Thank you

Answer 1

Pick e (think epsilon). You want to make n large enough so that abs(f_n - f) is smaller than e.

f is continuous on a closed and bounded set (compact) so you can reduce any cover to a finite subcover.

For each point in x in [a,b] pick a d-ball B(x,2*d) around x so that |f(x) - f(y)| < e for all y in B(x,2*d).

Now we consider the cover B(x, d) -- note the 2 is now gone.

Now pick a finite sub cover, so a FINITE set of points {x_i} such that B(x_i, d) covers [a,b] and f does not vary much in any of the balls. Let us order the x_i so they are increasing. Note that x_{i+1} - x_i < 2d or else this would not be a cover.

Now, pick N large enough so that |f_n(x_i) - f(x_i)| < e for all n >N.

Now let x be in [a,b]. x is in one of the balls, say B(t, d)

We want to show f(x) and f_n(x) are close. We are done of x is one of the x_i, so assume not and that x_i < x < x_{i+1}

Now, x is withing 2d of both x_i and x_{i+1} because those two points are at most 2d apart.

|f(x_i) - f(x)| < e
and
|f(x_{i+1}) - f(x)| < e
by contruction of the balls.

Also, because the f_n are monotonic,
f_n(x_i) < f_n(x) < f_n(x_{i+1})

Now
|f_n(x_i) - f_n(x_i)| < e
|f_n(x_{i+1}) - f_n(x_{i+1})| < e

|f_n(x) - f(x)| <= |f_n(x) - f_n(x_i)| + |f_n(x_i) - f(x_i)| + |f(x_i) - f(x)|

Now, term by term,
|f_n(x) - f_n(x_i)| < |f_n(x_{i+1}) - f_n(x_i)| because the fs are monotonic
And this term is
<= |f_n(x_{i+1}) - f(x_{i+1})| + |f(x_{i+1}) - f(x_i)| + |f(x_i) - f_n(x_i)|.

So substituting in the previous inequality we have
|f_n(x) - f(x)| < |f_n(x_{i+1}) - f(x_{i+1})| + |f(x_{i+1}) - f(x_i)| + |f(x_i) - f_n(x_i)| + |f_n(x_i) - f(x_i)| + |f(x_i) - f(x)|

Now,
|f_n(x_{i+1}) - f(x_{i+1})| < e because n is large
|f(x_{i+1}) - f(x_i)| < e because the points x_{i+1}, x_i are withing 2d of each other and by contruction the f values are withing e.
|f(x_i) - f_n(x_i)| < e because n is large.
|f_n(x_i) - f(x_i)| < e because n is large.
f(x_i) - f(x)| < e because x and x_i are close and contruction of the 2d balls.

So the expression is less than 5 e and hence we are done.

I think you can clean this up, but you get the idea.

To point out, the counter example below is not valid as it is on the open interval (0,1] and not the closed interval [0,1]. My proof uses the fact that the interval [a,b] is closed and therefore compact.

One more thing. In the proof above, you need to assume a = x_1 and b = x_n, which you can do.

Answer 2

The statement is correct and the convergence is indeed uniform. This result is known as Polya Theorem. Continuity of the f_n is not required.

The example ayayaya gave is not actually a counter example. His functions are defined on (0, 1], not [0, 1]. Since (0, 1] is not compact, the conditions required by Polya theorem aren't fully satisfied. And if he extends his definition to [0, 1], then his sequence will converge to the function given by f(0) = 0 and f(x) = 1 for x in (0, 1], which is discontinuous at x = 0. Again, one of the conditions required by Polya theorem is not satisfied.

So, we have no counter example.

I think Drdisk prof is OK, but maybe a simpler proof is possible.

Since f is continuous on the compact [a, b], it's uniformly continuous. So, given eps >0, there is d >0 such that |f(u) - f(v)| < eps for every u and v in [a, b] such that |u - v | < d. Take a partition of [a, b] with norm (length of the largest interval) is less than d.Let a = x_0, x_1...x_i,... x_m = b be the partition points. Since f_n --> f and the set {x_0, x_2...x_m} is finite, there exists k such that n >= k implies

|f_n(x_i) - f(x_i)| < eps for each i=0,1,...m. Therefore, for each i=0,1,...m-1

|f_n(x_(i+1) - f_n(x_i)| <= |f_n(x_(i+1) - f(x_(i+1))| + |f(x_(i+1)) - f(x_i)| + |f(x_i) - f_n(x_i)| < 3eps, in virtue of the previous inequality and of the uniform continuity of f on [a,b].


If x is in [a, b], then x is in one of the intervals [x_i, x_(i +1)], i=0,1,...m-1. Then, for n >=k,

|f_n(x) - f(x)| <= |f_n(x) - f_n(x_i)| + |f_n(x_i) - f(x_i)| + |f(x_i) - f(x)|

Since f_n is monotonic, |f_n(x) - f_n(x_i)| <= |f_n(x_(i+1) - f_n(x_i)| < 3 eps. So,

|f_n(x) - f(x)| < eps + 3ps + eps = 5eps, for every x in [a,b] and every n >= k. Since eps is arbitrary, it follows the convergence of f_n to f is uniform.

It's Interesting that it's not necessary that all of the f_n be increasing or decreasing, they can alternate. But if f is not constant, they will be eventually increasing or decreasing.

Answer 3

Our strategy is simple -- we need to find, for every ε>0, an N large enough so that ∀y∈[a, b], n>N ⇒ |f_n(y)-f(y)|<ε. We will start by using the continuity of f and the monotonicity of the f_n to show that around every point x there is a neighborhood over which we can choose such an N, then use the compactness of [a, b] to show that finitely many such neighborhoods cover [a, b], then finally choose an N that works for all of those finitely many neighborhoods (by choosing an N that works for each and then taking the maximum of such N).

Let ε>0. Select any point x∈[a, b]. Since f is continuous, there exists δ>0 such that whenever |y-x|<δ, |f(y)-f(x)|<ε/8. Choose c, d∈[a, b] such that x-δN₁, |f_n(c)-f(c)|<ε/8 and ∀n>N₂, |f_n(d)-f(d)|<ε/8. Now, let N = max (N₁, N₂). Then for any n>N, we have that |f_n(d)-f_n(c)| ≤ |f_n(d)-f(d)| + |f(d)-f(x)| + |f(x)-f(c)| + |f(c)-f_n(c)| < ε/8 + ε/8 + ε/8 + ε/8 = ε/2. Because the f_n are monotone, ∀n, ∀y∈[c, d], we have that |f_n(y)-f_n(c)| ≤ |f_n(d)-f_n(c)|. So for any n>N, y∈[c, d], we have |f_n(y)-f(y)| ≤ |f_n(y) - f_n(c)| + |f_n(c)-f(c)| + |f(c)-f(x)| + |f(x)-f(y)| ≤ ε/2 + ε/8 + ε/8 + ε/8 = 7ε/8 < ε. This establishes that for every x, there is an open (in [a, b]) set S containing x (specifically, S=(c, d) if x∈[a, b], S=[a, d) if x=a and S=(c, b] if x=b) for which there exists N such that whenever n>N, ∀y∈S⊆[c, d], |f_n(y)-f(y)| < ε.

Now, let C={S: S is open in [a, b] and ∃N s.t. (n>N ∧ y∈S) ⇒|f_n(y)-f(y)| < ε. Clearly, this a collection of open sets, and since ∀x, there exists such an open set containing x, C is in fact an open cover of [a, b]. [a, b] is compact, so C has a finite subcover, say, S₁, S₂... S_m. Then by the definition of C, ∃N₁, N₂... N_m such that (n>N_k ∧ y∈S_k) ⇒|f_n(y)-f(y)|<ε. So let N=max (N₁, N₂... N_m). Then suppose n>N. If y is any element of [a, b], then y must be an element of one of the S_k (as they cover [a, b]), and clearly, n>N_k, so |f_n(y)-f(y)|<ε. Therefore, we have found an N such that ∀y∈[a, b], n>N ⇒ |f_n(y)-f(y)|<ε.

As ε was chosen arbitrarily at the beginning of the problem, we can find such an N for any ε>0, thus we have ∀ε>0, ∃N s.t. ∀y∈[a, b], n>N ⇒ |f_n(y)-f(y)|<ε, which is to say the f_n converge uniformly on [a, b]. Q.E.D.

Answer 4

no longer lots artwork to coach. on account that s_n(x) converges uniformly to f(x) on [a,b] and [b,c], s_n(d) --> f(d) for all d in [a,b] union [b,c]. [a,b] union [b,c] is [a,c], so s_n(d) --> f(d) for all d in [a,c]. have been you meant to have some assumption approximately continuity?