Uniform Convergence of a Pathological Function

Question

Consider a function $f: [a_1, a_2]\to \mathbf{R}$ defined as the limit of the iterative sequence described below

The seed value of the iteration $f_0$ is the segment joining the points $(a_1, b_1)$ and $(a_2, b_2)$ We write that $$f_0 \equiv ((a_1, b_1), (a_2, b_2)).$$

We now define a recursive function $F$ which maps a segment to a collection of three segments. $F: ((a_1, b_1), (a_2, b_2)) \mapsto((a_1, b_1), (\frac{2a_1+a_2}{3}, \frac{b_1+2b_2}{3}), (\frac{a_1+2a_2}{3}, \frac{2b_1+b_2}{3}), (a_2, b_2))$ Using this recursion, we define the next function $f_1$ as follows: $$f_1 := F(f_0) \equiv ((a_1, b_1) (\frac{2a_1+a_2}{3}, \frac{b_1+2b_2}{3}), (\frac{a_1+2a_2}{3}, \frac{2b_1+b_2}{3}), (a_2, b_2))$$ $f_1$ looks as follows:

We can now form a sequence $(f_n)_{n\geq0}$ wherein we apply the function $F$ on each of the segments of $f_{n-1}$ to get $f_{n}.$

Therefore, $f_n$ would be a collection of $3ˆ{n}$ segments and would be represented as a $(3ˆ{n}+1)-$tuple of points. Each of these functions is kind of like a zigzag path. I call each vertex of this path a "Sharp Point"

Situation after a few iterations ($f_3$ and $f_4$ are shown) looks like:

I have to prove that the limit of this sequence uniformly converges

My attempt:

I am trying to prove that the sequence is Cauchy.

Without loss of generality, let $b_1 < b_2$ and $m < n$

It is clear that the iteration does not change the end points of the segments. Therefore, $$f_m(x) = f_n(x) \implies x = a_1 + \frac{p}{3ˆm}, p \in \mathbf{N}, 0 < p < 3ˆm$$ Therefore, all sharp points of $f_m$ are sharp points of $f_n$ as well.

Using this fact and the fact that $f_m$ and $f_n$ are continuous we can see that $f_m(x) - f_n(x)$ would maximise at those sharp points of $f_n$ that are not sharp points of $f_m$.

Now, $||f_n - f_m||_{[a_1, a_2]} = sup[f_m(x) - f_n(x)] = max[f_m(s) - f_n(s)]$ where s is some sharp point of $f_n$ that is not a sharp point of $f_m$

I am unsure how to proceed further. Any help would be greatly appreciated. If there is a better approach then that would be great as well.

Thanks a lot.

Answer 1

The "fundamental operation" in passing from $f_{n}$ to $f_{n+1}$ replaces a segment by a zig-zag of three segments.

On each segment $[a_{1}, a_{2}]$ where $f_{n}$ is affine and changes by $|f_{n}(a_{2}) - f_{n}(a_{1})| = |b_{2} - b_{1}| = \Delta y$, we have $$ \max_{a_{1} \leq x \leq a_{2}} |f_{n+1}(x) - f_{n}(x)| = \tfrac{1}{3}\Delta y. $$ If $\Delta y_{n}$ denotes the maximum absolute change of an affine portion of $f_{n}$ (taken over all segments), then:

• $\Delta y_{n+1} \leq \frac{2}{3}\, \Delta y_{n}$, so $\Delta y_{n} \leq (\frac{2}{3})^{n}\, \Delta y_{0}$ by induction;
• $\|f_{n+1} - f_{n}\|_{\infty} \leq \frac{1}{3}\, \Delta y_{n} \leq \frac{1}{3} \cdot \frac{2^{n}}{3^{n}}\, \Delta y_{0}$.

The triangle inequality and summing a geometric series show that for all integers $n \geq 0$ and $m \geq 1$, \begin{align*} \|f_{n+m} - f_{n}\|_{\infty} &\leq \sum_{k=0}^{m-1} \|f_{n+k+1} - f_{n+k}\|_{\infty} \\ &\leq \frac{1}{3} \sum_{k=0}^{m-1} \frac{2^{n+k}}{3^{n+k}}\, \Delta y_{0} \\ &< \frac{1}{3} \sum_{k=0}^{\infty} \frac{2^{n+k}}{3^{n+k}}\, \Delta y_{0} \\ &= \frac{2^{n}}{3^{n}}\, \Delta y_{0} \end{align*} independently of $m$. That is, the sequence $(f_{n})$ is uniformly Cauchy, so it converges uniformly to a function $f$.