# Uniform convergence and lengths

Consider the following sequence of piece-wise smooth functions $f_n:[0,2]\rightarrow \mathbb{R}$ (below drawn first two functions by their graphs, and is contunued then in natural way.)

The function $f:[0,2]\rightarrow \mathbb{R}$, $f(x)=0$ for all $x$ is the limit of this sequence.

Q. 1 Note that the lengths of graph of each $f_n$ is $2\sqrt{2}$, but the length of limiting function is not $2\sqrt{2}$; why this happens? (Such problem appears as a puzzle shown below picture in some book, but I was looking it through a sequence of functions, it convergence etc.)

Here sequence $f_n$ converges to $f$ uniformly; but still the length of their graphs do not converges to length of graph of limit of $f_n$. This raises following question:

Q If $f_n:[a,b]\rightarrow \mathbb{R}$ is a sequence of piecewise smmoth functions, and converging uniformly to $f:[a,b]\rightarrow \mathbb{R}$, then under what more conditions on $f_n$, we can guarantee the convergence of lengths of $f_n$ to length of $f$?

Let $$\mathscr{F}= \{\,f_\alpha\}_{\alpha \in A}$$ be a family of real-valued functions defined on $$\left[a,b\right]$$. We say that $$\mathscr{F}= \{\,f_\alpha\}_{\alpha \in A}$$ is absolutely equicontinuous on $$\left[a,b\right]$$ if and only if for every $$\varepsilon>0$$ there exists $$\delta>0$$ so that $$$$\sum_{k \mathop = 1}^N \left|\, f_\alpha(b_k)-f_\alpha(a_k)\right|<\varepsilon \; \text{ for all } \alpha \in A \, \text{ whenever }\, \sum_{k \mathop = 1}^N \left(b_k-a_k\right)<\delta ,$$$$ and the intervals $$\left(a_k,b_k \right)$$, $$\, k=1, \ldots, N$$ are disjoint subsets of $$\left[a,b\right]$$.

Let $$\{\,f_n\}_{n=1}^\infty$$ and $$\{g_n\}_{n=1}^\infty$$ be the sequences of real-valued functions on $$\left[0,2\right]$$ such that given $$n \in \mathbb{N}$$, $$\; f_n$$ and $$g_n$$ are respectively defined by

\begin{align}f_n(x) := \begin{cases} x-\frac{2k-2}{n} & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right], \\ \frac{2k}{n}-x& \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] \end{cases} && (k=1,\ldots,n) \end{align}

$$$$

\begin{align}g_n(x) := \begin{cases} n^{n-1}\left(x-\frac{2k-2}{n}\right)^n & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right], \\ n^{n-1}\left(\frac{2k}{n}-x\right)^n & \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] \end{cases} && (k=1,\ldots,n) \end{align}

(i) The family of functions $$\mathscr{F}:=\{\,f_n\}_{n \in \mathbb{N}}$$ is absolutely equicontinuous on $$\left[0,2\right]$$.

Let $$\varepsilon>0$$ be given. Choose $$\delta = \varepsilon$$. Since $$\left|\,f_n(d)-f_n(c) \right| \leq (d-c)$$ for all $$n \in \mathbb{N}$$ and $$\left(c,d \right) \subseteq \left[0,2\right]$$, it follows that $$\mathscr{F}$$ is an absolutely equicontinuous family on $$\left[0,2\right]$$.

(ii) The family of functions $$\mathscr{G}:=\{g_n\}_{n \in \mathbb{N}}$$ is NOT absolutely equicontinuous on $$\left[0,2\right]$$.

Take $$\varepsilon=\frac{1}{2}(1-e^{-1})$$, and let $$\delta>0$$ be given. By the Archimedean Principle we may find a positive integer $$N$$ so that $$N > \frac{1}{\delta}$$. Now given a positive integer $$n \geq N$$, we set $$a_{n,k}:=\left(\frac{2k-1}{n} - \frac{1}{n^2} \right)$$ and $$b_{n,k}:=\frac{2k-1}{n}$$ $$\, \left(k=1, \ldots,n \right)$$. So for example we have that \begin{aligned} &\sum_{k \mathop = 1}^N \left(b_{N,k}-a_{N,k}\right)=\frac{1}{N}<\delta \;\, \text{, and} \\& \sum_{k \mathop = 1}^N \left|\, g_N(b_{N,k})-g_N(a_{N,k})\right|= 1-\left(1-\frac{1}{N}\right)^N . \end{aligned} Since $$\lim_{N \to \infty} \left[1-\left(1-\frac{1}{N}\right)^N \right]=1-e^{-1}$$, it follows that for sufficiently large $$n \geq N$$ we shall have \begin{aligned} &\sum_{k \mathop = 1}^n \left(b_{n,k}-a_{n,k}\right)=\frac{1}{n}<\delta \;\, \text{, BUT} \\& \sum_{k \mathop = 1}^n \left|\, g_n(b_{n,k})-g_n(a_{n,k})\right|> \varepsilon . \end{aligned}

So $$\mathscr{G}$$ is not an absolutely equicontinuous family on $$[0,2]$$.

Note: $$\mathscr{G}$$ is a uniformly equicontinuous family on $$[0,2]$$ because given $$n \in \mathbb{N}$$ we have that $$|g_n(x)|\leq \frac{1}{n} = h_n(x)$$ for all $$x \in [0,2]$$. To be clear, this implies that $$\{g_n\}_{n=1}^\infty$$ converges uniformly to $$g \equiv 0$$ on $$[0,2]$$, since the sequence of dominating functions $$\{h_n\}_{n=1}^\infty$$ converges uniformly to $$0$$. Let $$\varepsilon>0$$ be given. Since $$g_n \to 0$$ uniformly on $$[0,2]$$, we know there is a positive integer $$N$$ so that $$|g_n(x)|<\frac{\varepsilon}{2}$$ for all $$x \in [0,2]$$ whenever $$n \geq N$$. Hence $$|g_n(x)-g_n(y)| \leq |g_n(x)|+|g_n(y)|<\varepsilon$$ whenever $$n \geq N$$, and $$x,y \in [0,2]$$. So we have the $$N-1$$ functions $$\{g_1, \ldots, g_{N-1} \}$$ left to find a uniform delta for that will work for every function in the family $$\mathscr G$$. We know every function in the finite collection $$\{g_1, \ldots, g_{N-1} \}$$ is continuous on $$[0,2]$$, as each of these functions is piecewise continuous with pieces that agree at each endpoint (Pasting/Gluing Lemma). Hence each of these functions is uniformly continuous on $$[0,2]$$, by: Proof verification for a couple of theorems regarding Lebesgue's number. Thus for each $$k=1, \ldots, N-1 \,$$ we may find a positive number $$\delta_k$$ so that $$|g_k(x)-g_k(y)|<\varepsilon$$ whenever $$|x-y|<\delta_k$$, and $$x,y \in [0,2]$$. Setting $$\delta = \min \{\delta_1, \ldots, \delta_{N-1} \}$$ shows that $$\mathscr G$$ satisfies the definition of a uniformly equicontinuous family on $$[0,2]$$.

Let $L(\,f)$ denote the length of the graph \begin{align} \Gamma_f = \{t+if(t) \in \mathbb{C} : t \in [0, 2]\} \,. \end{align}

Let $\{\,f_n\}_{n=1}^\infty$ be a sequence of real-valued continuous functions that converge uniformly on $[0, 2]$ to the function $f:=\lim\limits_{n \to \infty}\,f_n$. We have that $\,L(\,f) = \lim\limits_{n \to \infty} L(\,f_n)$ holds if and only if the sequence of functions $\{\,f_n\}_{n=1}^\infty$ forms on $[0,2]$ an absolutely equicontinuous family such that $f^{'}_n$ converges in measure to $f'$.

I can't give a formal proof of the preceding statement. Though I took the liberty of coding up a worse counterexample sequence $\{g_n\}_{n=1}^\infty$ that I found more insightful \begin{align}f_n(x) := \begin{cases} x-\frac{2k-2}{n} & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right], \\ \frac{2k}{n}-x& \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] \end{cases} && (k=1,\ldots,n) \end{align}



\begin{align}g_n(x) := \begin{cases} n^{n-1}\left(x-\frac{2k-2}{n}\right)^n & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right], \\ n^{n-1}\left(\frac{2k}{n}-x\right)^n & \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] \end{cases} && (k=1,\ldots,n) \end{align}

\begin{aligned} & L(g_j) =2j \int_0^{1/j} (1+j^{2j}t^{2j-2})^{1/2} dt \\& L(g_j) \leq L(g_{j+1})\end{aligned} \: \;\: \;(\,j=1,2,\ldots)

Note: $g_1(x)=f_1(x)$ for all $x \in [0,2].$

I should mention that there is a sequence of functions similar to $\{\,f_n\}_{n=1}^\infty$ $(*)$ described in a comment by ThePortakal, here: Sequence of functions that converges uniformly in $R$, derivatives does not converge punctually at any point of $R$.

$(*)$ The family of functions described by ThePortakal is absolutely equicontinuous on $[0,2]$, and the graphs corresponding to the functions in this family will have constant length $\pi$ on $[0,2]$.

I have decided to add a picture of the work that went into my original answer (excuse the notation abuse with $\,f_n$ on the right-hand side)

Along with Beginner's interesting question here and ThePortakal's comment in another thread, this Dover book was helpful in motivating the sequence $\{g_n\}_{n=1}^\infty$.

• I should say that I currently feel dubious about these statements concerning "$L(\,f) = \lim\limits_{n \to \infty} L(\,f_n)$". I read about these conditions somewhere when I originally came across this question. I have lost the reference, and I have never taken a course in measure theory. I do know a bit of mathematical analysis not including measure theory, so this is what I may speak from with confidence. Anyways, hopefully I haven't made your question about absolute equicontinuity out of nothing. I would upvote your question a second time if I could, because I did find it quite riveting. May 7, 2018 at 21:46
• I can say that $\lim\limits_{n \to \infty}\mu ( \{x \in [0,2] : |f'(x)-f_n'(x)| \geq \frac{1}{2} \})=2$, since $|f_n'(x)|=1$ for all $n \in \mathbb{N}$ and $x \in [0,2]$, and $f_n \to 0$ uniformly on $[0,2]$ (like $g_n$ and $h_n$). I would just feel more comfortable if I still remembered the theorem that I ultimately referenced to make sure I stated its conditions/assumptions in my original answer correctly. I do believe the reference where I looked the theorem up used different terminology than "absolutely equicontinuous" but I prefer this terminology for the given definition. May 8, 2018 at 2:04

Suppose $$\{f_n\}_{n=1}^\infty$$ is a sequence of real-valued functions such that $$\lim_{n \to \infty} f_n(a)$$ exists, $$f_n'$$ converges in measure to $$g$$ on $$\mathbb R$$, $$|g| \in L^1([a,b]),$$ and $$\{|f_n'|\}_{n=1}^\infty \subset L^1([a,b])$$ forms a uniformly integrable (or rather, $$\{f_n\}_{n=1}^\infty$$ forms an absolutely equicontinuous) family. Let $$f(x):=\lim_{n \to \infty} f_n(a)+\int_a^x g dm$$. Let $$\varepsilon>0$$ be given. Since $$\{|g|\} \cup \{|f_n'|\}_{n \in \mathbb N}$$ is a uniformly integrable family on $$[a,b]$$ (when applying the definition of uniformly integrable, we may take $$\delta=\min \{\delta_1, \delta_2\}$$), there is $$\delta>0$$ so that $$\int_E |f_n'|dm<\varepsilon$$ for all $$n \in \mathbb N$$ and $$\int_E |g|dm<\varepsilon$$ whenever $$m(E)<\delta$$ and $$E \subset [a,b]$$. Given $$n \in \mathbb N$$, let $$E_n:=\{x \in [a,b]:|f_n'(x)-g(x)|\geq \varepsilon\}$$. Since $$f_n' \to g$$ in measure, there is $$N \in \mathbb N$$ so that $$m(E_n)<\delta$$ whenever $$n \geq N.$$ So we have $$n \geq N$$ implies that \begin{align*} \int_{[a,b]} |f_n'-g|dm &\leq \int_{[a,b] \setminus E_n} |f_n'-g|dm + \int_{E_n} |f_n'-g|dm \\&\leq \int_{[a,b] \setminus E_n} |f_n'-g|dm + \int_{E_n} |f_n'|dm +\int_{E_n} |g|dm \\& < (b-a)\varepsilon + \varepsilon + \varepsilon= (b-a+2)\varepsilon. \end{align*} So $$\lim_{n \to \infty} \int_{[a,b]} |f_n'-f'|dm =\lim_{n \to \infty} \int_{[a,b]} |f_n'-g|dm =0$$, and $$f_n$$ converges pointwise (actually uniformly* - this observation is just made so we may understandably write $$L(f)$$ instead of $$L(\lim_{n \to \infty} f_n)$$) to $$f$$ on $$[a,b]$$. Since $$\left| \int_{[a,b]} \sqrt{1+{f_n'}^2} dm - \int_{[a,b]} \sqrt{1+{f'}^2} dm\right| \le \int_{[a,b]} |f_n'-f'| dm,$$ we have that $$\lim_{n \to \infty} L(f_n) = L(f)$$.

*Suppose $$\{f_n\}_{n \in \mathbb N}$$ is a uniformly equicontinuous family of real-valued functions on $$[a,b]$$, $$f$$ is continuous on $$[a,b]$$, and $$f_n \to f$$ pointwise on $$[a,b]$$ (technically we could dispose of the requirement that $$f$$ is continuous and only require $$\{f_n\}_{n \in \mathbb N}$$ to be equicontinuous on $$[a,b]$$ and still have the same result but the chosen assumptions will already be true in the context above). Let $$\varepsilon>0$$ be given. We may find $$\delta_1>0$$ and $$\delta_2>0$$ so that $$|f_n(x)-f_n(y)|<\varepsilon$$ for all $$n \in \mathbb N$$ and $$|f(x)-f(y)|<\varepsilon$$ whenever $$|x-y|<\delta=\min\{\delta_1,\delta_2\}$$ and $$x,y \in [a,b]$$. By the Archimedean principle, we may find $$M \in \mathbb N$$ so that $$\frac{b-a}M<\delta$$. Given $$k \in \{1,\ldots, 2M-1\}$$, let $$x_k:=a+k\frac{b-a}{2M}$$. Given $$k \in \{1,\ldots, 2M-1\}$$, we may find $$N_k \in \mathbb N$$ so that $$|f_n(x_k)-f(x_k)|<\varepsilon$$ whenever $$n \geq N_k$$ ($$f_n \to f$$ pointwise). Take $$N=\max\{N_1, \ldots, N_{2M-1}\}$$. Let $$x \in [a,b]$$. So $$|x-x_k|<\delta$$ for some $$k \in \{1,\ldots, 2M-1\}$$, and we have $$|f_n(x)-f(x)|\leq |f_n(x)-f_n(x_k)|+|f_n(x_k)-f(x_k)|+|f(x_k)-f(x)|<3\varepsilon$$ whenever $$n \geq N$$. So $$\sup_{x \in [a,b]} |f_n(x)-f(x)| \leq 3\varepsilon$$ whenever $$n \geq N$$. Hence $$\lim_{n \to \infty} \sup_{x \in [a,b]} |f_n(x)-f(x)|=0$$. In other words, $$f_n \to f$$ uniformly on $$[a,b]$$.

Considering the Vitali convergence theorem, these assumptions turn out to be equivalent to saying $$f_n' \to f'$$ in $$L^1([a,b])$$** which in some sense seems underwhelming. And since we are dealing with spaces of finite measure, $$f_n' \to f'$$ uniformly would also imply that $$\lim_{n \to \infty} L(f_n) = L(f)$$.

**Suppose $$\{f_n\}_{n=1}^\infty \subset L^1$$, $$f \in L^1$$, and $$\lim_{n \to \infty} \int \left|f_n-f \right|dm =0$$. Let $$\varepsilon>0$$ be given.

By Corollary $$3.6$$ (of Folland - what the corollary says should be clear from the rest of this sentence), we may find $$\delta'>0$$ so that $$\left| \int_E f d m\right|< \varepsilon$$ whenever $$m(E)<\delta'$$. Since $$\lim_{n \to \infty} \int \left|f_n-f \right|dm =0$$, there is $$N \in \mathbb N$$ so that $$\int \left|f_n-f \right|dm<\varepsilon$$ whenever $$n \geq N$$. So if $$n \geq N$$, then we have \begin{align*} \left| \int_E f_n d m \right| &\leq \left| \int_E f_n-f d m\right|+\left| \int_E f d m \right| \\&\leq \int_E \left|f_n-f \right|dm +\left| \int_E f d m\right| \\&\leq \int \left|f_n-f \right|dm +\left| \int_E f d m\right| \\&<\varepsilon+\varepsilon=2\varepsilon. \end{align*} Given $$j \in \{1, \ldots, N-1\}$$, we may find $$\delta_j>0$$ so that $$\left| \int_E f_j d m \right|< \varepsilon$$ whenever $$m(E)<\delta_j$$ (Corollary $$3.6$$). Let $$\delta''=\min\{\delta_1, \ldots, \delta_{N-1}\}$$. Thus $$\left| \int_E f_j d m\right|< \varepsilon$$ for all $$j \in \{1, \ldots, N-1\}$$ whenever $$m(E)<\delta''.$$ Let $$\delta=\min \{\delta', \delta''\}$$. Thus $$\left| \int_E f_n d m\right|< \varepsilon$$ for all $$n \in \mathbb N$$ whenever $$m(E)<\delta.$$ In other words, $$\{f_n\}_{n \in \mathbb N}$$ is uniformly integrable.

Let $$E_n=\{x: |f_n(x)-f(x)|\geq \varepsilon \}$$. Since $$\lim_{n \to \infty} \int \left|f_n-f \right|dm =0$$, there is $$N' \in \mathbb N$$ so that $$\int \left|f_n-f \right|dm<\varepsilon^2$$ whenever $$n \geq N'$$. So if $$n \geq N'$$, then we have $$\varepsilon m(E_n) \leq \int_{E_n} \left|f_n-f \right|dm \leq \int \left|f_n-f \right|dm <\varepsilon^2.$$ In other words, $$f_n \to f$$ in measure.