# Sergei Yakovenko's blog: on Math and Teaching

## Continuation of continuity

Let $f\colon [0,1]\times[\alpha,\beta]\to\mathbb R^2$ be a continuous function defined on the closed rectangle $[0,1]\times[\alpha,\beta]$. Then the functions $f_a(x)=f(x,a)$ are all defined on the segment $[0,1]$ and “converge” to $f_{a_*}(x)$ as $a\to a_*\in[\alpha,\beta]$. If $a_1,a_2,\dots, a_n,\dots$ is a sequence of values of $a$ converging to $a_*$, $\lim a_n=a_*$, then we have a sequence of functions $f_n=f_{a_n}\colon[-1,1]\to\mathbb R$, all defined on the same segment, which also “converges” to the limit $f_*=f_{a_*}$

The meaning of the “convergence of functions” is not yet defined.

Problem. Prove that in the above notations, for any point $x\in[0,1]$ the sequence of numbers $f_n(x)$ converges to the number $f_*(x)$. This type of convergence is called pointwise convergence (התכנסות נקודתית).

Example. Prove that the functions $f_n(x)=x^n,~n=1,2,\dots$, converge pointwise to a certain limit function $f_*\colon[0,1]\to\mathbb R$. Find this function.

Example. Let $\phi(x)=\frac1{1+x^2}$ be the function defined on the entire axis $\mathbb R$, and denote $f_n(x)=\phi(x-n)\colon\mathbb R\to\mathbb R$ (translation to the right by $n$). The same question: does a limit exist?

Formally, the pointwise convergence is defined by the following formula: $\lim f_n=f_*$, if $\forall x\in [0,1],~\forall \varepsilon >0,~\exists N=N(x,\varepsilon),~\forall n\ge N,~|f_n(x)-f_*(x)|<\varepsilon$. Clearly, this is the same as the formula

$\forall \varepsilon >0,~\forall x\in [0,1],~\exists N=N(x,\varepsilon),~\forall n\ge N,~|f_n(x)-f_*(x)|<\varepsilon.$

However, the “convergence” which comes from the continuity in two variables, is stronger.

Theorem. If $f(x,y)$ is a continuous function on the rectangle, as above, then for any $\varepsilon>0$ there exists a positive $\delta>0$ such that for any $a$ with $|a-a_*|<\delta$, the functions $f_a$ and $f_{a_*}$ differ at most by $\varepsilon$ on the entire segment $[0,1]$: $\forall x\in[0,1],~|f_a(x)-f_{a_*}(x)|<\varepsilon$.

Corollary. In the above notation, $f_n$ converges to $f_*$ “simultaneously” (uniformly, במידה שווה):

$\forall\varepsilon>0,~\exists N=N(\varepsilon):~\forall x\in[0,1],~\forall n\ge N,~|f_n(x)-f_*(x)|<\varepsilon.$

### Functions as “points” (vectors)

Continuous functions defined on the same interval, can be added, subtracted, multiplied by constants etc, which makes them similar to vectors in $\mathbb R^3$. We can also define the “distance” between the functions as follows, $d(f_1,f_2)=\max_{x\in[0,1]}|f_1(x)-f_2(x)|$. This number is always non-negative and finite:

1. $d(f_1,f_2)=0\iff f_1=f_2$;
2. $d(f_1,f_2)=d(f_2,f_1)$ (symmetry);
3. $d(f_1,f_3)\le d(f_1,f_2)+d(f_2,f_3)$ (the triangle inequality).

Proposition/Definition. The sequence of functions $\{f_n\}$ converges uniformly to $f_*$, if $d(f_n,f_*)\to0$.

Question. In which of the examples above the convergence is uniform?

### Can a limit of continuous functions be disconjugate?

A sequence of pointwise converging continuous functions may converge to a discontinuous function (see the first example above).

Theorem. The limit of uniformly converging on a closed sequence continuous functions is again continuous.

Example. The snowflake is a continuous closed curve of “infinite length”.

### Peano (plane-filling) curve

This is a continuous map $\gamma\colon [0,1]\to \mathbb R^2$ which fills completely (i.e., passes through each point of) the unit square. There are many examples of such curves, below is the example constructed by D. Hilbert. Giuseppe Peano was an Italian mathematician

### Sketch of the justification

In both cases the “malicious” curve is obtained as the limit of an infinite sequence of planar polygonal (hence continuous) curves $\gamma_n\colon [0,1]\to\mathbb R^2$. By construction, each next curve $\gamma_{n+1}$ differs from its predecessor $\gamma_n$ by an explicitly limited “modification”. In the example with the Peano curve, we subdivide the initial unit square into for small squares of the “first level” and draw the curve crossing each of the four squares in the specified order. Then each 1st level square is again subdivided into 4 tiny sub-squares of the “second level” and each segment of the curve is modified so as it (1) remains inside the same 1st level square, yet (2) crosses each of the four 2nd level squares as plotted. Hence the curve $\gamma_2$ differs from  the curve $\gamma_2$ by no more than $\sqrt 2$. In the same way $\|\gamma_3-\gamma_2\|\le \sqrt 2/2$$\|\gamma_4-\gamma_3\|\le \sqrt 2/2^2$ etc.

Thus it is clear that for any $t\in[0,1]$ the limit $\lim_{n\to\infty}\gamma_n(t)$ exists, and $|\gamma_n(t)-\gamma_*(t)|\le \sqrt{2}/2^{n-3}$. Thus the convergence of the functions is uniform, and therefore the limit is also a continuous function.

How to show that it indeed passes through each point of the unit square? Note that the curve $\gamma_n$ passes through centers of all squares of the $n$th level. Any point $a_*$ of the initial square can be approximated by centers of the squares of sufficiently high level. There exists an infinite sequence of points $a_n\in\gamma_n([0,1])$  which converges to $a_*$, i.e., the infinite sequence of moments $t_n\in[0,1]$ such that $\lim_{n\to\infty}\gamma_n(t_n)=a$. Passing to a subsequence if necessary, we can assume (since the segment [0,1] is compact) that the sequence $\{t_n\}$ converges to a limit $t_*$. One can easily check that because of the uniform continuity, $\gamma_*(t_*)=a_*$.

Problem. Write the detailed accurate exposition of this proof.