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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.)

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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 $f$?

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3 Answers 3

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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 \begin{equation} \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 , \end{equation} 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{equation} \end{equation}

\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 $[0,2]$.

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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{equation} \end{equation}

\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{equation} \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) \end{equation}

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 $R$.

enter image description here

$(*)$ 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) enter image description here


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

enter image description here

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    $\begingroup$ 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. $\endgroup$
    – M A Pelto
    May 7, 2018 at 21:46
  • $\begingroup$ 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. $\endgroup$
    – M A Pelto
    May 8, 2018 at 2:04
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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.

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