Sequence of continuous functions convergent to an increasing real function

[Arnold1]

Hi. Could help me with the following problem?

Let $$f$$ be a real function, increasing on $$[0,1]$$.

Does there exists a sequence of functions, continuous on $$[0,1]$$, convergent pointwise to $$f$$? If so, how to prove it?

I would really appreciate any help.

Thank you.

 

[chisigma]

Hi. Could help me with the following problem?

Let $$f$$ be a real function, increasing on $$[0,1]$$.

Does there exists a sequence of functions, continuous on $$[0,1]$$, convergent pointwise to $$f$$? If so, how to prove it?

I would really appreciate any help.

Thank you.

If f(*) is continouos in [0,1] then the sequence... $\displaystyle f_{n} (x) = \frac{a_{0}}{2} + \sum_{k=1}^{n} (a_{n}\ \cos 2\ k\ \pi\ x + b_{n}\ \sin 2\ k\ \pi\ x)$ (1)... where $a_{k}$ and $b_{k}$ are the Fourier coefficients of f(*), pointwise converges to f(*) in (0,1)... Kind regards $\chi$ $\sigma$

 

[tarantella]

We can use a convolution with a sequence $\{K_n\}$ of regularizing kernels, that is, not negative continuous functions of integral $1$.

 

[I like Serena]

Hi. Could help me with the following problem?

Let $$f$$ be a real function, increasing on $$[0,1]$$.

Does there exists a sequence of functions, continuous on $$[0,1]$$, convergent pointwise to $$f$$? If so, how to prove it?

I would really appreciate any help.

Thank you.

Welcome to MHB, Arnold! :)

Suppose your function f has a jump in it.
That is, suppose it is not continuous.
Then no sequence of continuous functions can converge to it.

Formally you need an ($\varepsilon, \delta$)-proof that says the same thing.

 

[Opalg]

Let $$f$$ be a real function, increasing on $$[0,1]$$.

Does there exists a sequence of functions, continuous on $$[0,1]$$, convergent pointwise to $$f$$? If so, how to prove it?

This seems like a difficult problem. An increasing function on $[0,1]$ is continuous everywhere except at an at most countable number of points $x_k$ where there is a jump discontinuity. At each of these points, the left and right limits \(\displaystyle f(x_k-) = \lim_{x\nearrow x_k}f(x)\) and \(\displaystyle f(x_k+) = \lim_{x\searrow x_k}f(x)\) will exist (where for completeness you should define \(\displaystyle f(0-) = f(0)\) and \(\displaystyle f(1+) = f(1)\)), and the value of $f$ at the point $x_k$ itself could be anywhere in the interval $[f(x_k-),f(x_k+)]$. If you want to find a sequence of functions $f_n$ which converge to $f$ at every point of the interval, then you need to ensure that $f_n(x_k) \to f(x_k)$ at each point of discontinuity.

The drawback to the sort of regularising functions proposed by girdav is that these will treat each jump point in the same way. They can be tailored to converge to the midpoint of the jump, or the bottom point, or the top point, or indeed to any given fraction of the jump interval. But if the positions of $f(x_k)$ in their jump intervals vary, you need to tailor your approximating functions individually to each jump point, to get them to converge to the right place in the jump interval. Intuitively, this should certainly be possible, but I don't see a convenient way to achieve it.

 

[Opalg]

See if this construction works (I think it does).

Let $\{x_k\}$ be an enumeration of the points where $f$ has a jump discontinuity. For each $k$, let $\delta_k = f(x_k) - \frac12\bigl(f(x_k+) + f(x_k-)\bigr)$, so that $\delta_k$ is the amount by which $f(x_k)$ deviates from the midpoint of the jump. Notice that $\sum|\delta_k|$ converges, because that sum must be less than $f(1)-f(0)$ (the total variation of $f$ over the interval).

For $n=1,\ 2,\ 3,\ \ldots$, define \(\displaystyle g_n(x) = \frac n2\int_{x-(1/n)}^{x+(1/n)}f(t)\,dt\) and \(\displaystyle h_n(x) = \sum_{\{k\;:\;x-(1/n) \leqslant x_k \leqslant x+(1/n)\}}\bigl(1- n|x-x_k|\bigr)\delta_k\), and let $f_n(x) = g_n(x) + h_n(x).$ (To cope with integrals that extend beyond the ends of the unit interval, define $f(t) = f(0)$ when $t<0$, and $f(t) = f(1)$ when $t>1$.)

Then $g_n$ is continuous (because it is an integral), $h_n$ is continuous (because it is a uniformly convergent sum of continuous functions), and therefore $f_n$ is continuous. The big question is whether $f_n$ converges pointwise to $f$. The reason I think it does is that $g_n(x)$ represents the mean value of $f$ in the interval $[x-(1/n), \,x+(1/n)]$. As $n\to\infty$, that will converge to $f(x)$ at all points of continuity. At a jump, it will converge to the midpoint of the jump. As for $h_n(x)$, if $f$ is continuous at $x$ then for $n$ large enough there will be no large jumps in the interval $[x-(1/n), \,x+(1/n)]$, and therefore $h_n(x)$ will be small. On the other hand, if $x=x_k$ for some $k$ then, again for $n$ large enough, the sum of all the other jumps in the interval $[x_k-(1/n), \,x_k+(1/n)]$ will be much smaller than the jump at $x_k$, and so $h_n(x_k)$ will be close to $\delta_k$. Thus $f_n(x_k)$ will be close to $\frac12\bigl(f(x_k+) + f(x_k-)\bigr) + \delta_k = f(x_k).$

I don't have the energy to put all the $\varepsilon$s and $\delta$s into that argument to make it stand up properly, but I hope that it points in the right direction.

 

[chisigma]

Let suppose to have a finite set of $n_{j}$ jump discontinuities of f(*) in [0,1] choose $n \ge n_{j} + 2$ distinct points in [0,1] with the following criterion...

a) $x_{0} = 0,\ x_{n}=1$

b) $x_{k+1} > x_{k}\ k=0, 1, ..., n $

c) all the points of jump discontinuity belong to the set of $x_{k}$

d) $\lim_{n \rightarrow \infty} \text{max} (x_{k+1}-x_{k})=0$

The we define the set of functions...

$\displaystyle u_{k} (x) = \begin{cases} 0 & \text{if}\ x< x_{k}\\ \frac{f(x_{k+1})-f(x_{k})}{x_{k+1}-x_{k}}\ (x-x_{k})\ \text {if}\ x_{k} \le x \le x_{k+1}\\ f(x_{k+1})- f(x_{k}) &\text{if}\ x> x_{k+1}\end{cases}$ (1)

Under these hypotheses a set of functions pointwise converging to f(*) should be...

$\displaystyle f_{n}(x)= \sum_{k=0}^{n-1} u_{k} (x)$ (2)

Kind regards

$\chi$ $\sigma$

 

[Arnold1]

@chisigma Maybe it'a a stupid question, but I am not sure about one thing. We choose n points, right? From condition 1) we get n+1 points, from 2) we get \(\displaystyle x_{n+1}>x_n\) and \(\displaystyle x_1>x_0\). Am I missing something? And could you tell me how to check that \(\displaystyle f_n(x)\) is pointwise convergent to \(\displaystyle f\)?

Does it suffice to show that \(\displaystyle f_n(x_k) \rightarrow f(x_k) \ \ (n \rightarrow + \infty)\)?

Here is my attempt:

\(\displaystyle \lim _{x \rightarrow x_k-} u_k(x)=0\)

\(\displaystyle f(x_k)=f(x_{k+1})=0\) from 4)

\(\displaystyle \lim _{x \rightarrow x_k+} u_k(x)=0\) from 4) and continuity of \(\displaystyle f\) between the points from \(\displaystyle \{x_k\} _{ k\in \{0,1,...,n\}}\)

 

[tarantella]

http://www.whitman.edu/mathematics/.../2007/huh.pdf contains some stuff about Baire class 1 functions, that is, pointwise limits of sequences continuous functions.
There are criteria to determine whether a function is Baire class 1 or not.

 

[chisigma]

@chisigma Maybe it'a a stupid question, but I am not sure about one thing. We choose n points, right? From condition 1) we get n+1 points, from 2) we get \(\displaystyle x_{n+1}>x_n\) and \(\displaystyle x_1>x_0\). Am I missing something? And could you tell me how to check that \(\displaystyle f_n(x)\) is pointwise convergent to \(\displaystyle f\)?

Does it suffice to show that \(\displaystyle f_n(x_k) \rightarrow f(x_k) \ \ (n \rightarrow + \infty)\)?

Here is my attempt:

\(\displaystyle \lim _{x \rightarrow x_k-} u_k(x)=0\)

\(\displaystyle f(x_k)=f(x_{k+1})=0\) from 4)

\(\displaystyle \lim _{x \rightarrow x_k+} u_k(x)=0\) from 4) and continuity of \(\displaystyle f\) between the points from \(\displaystyle \{x_k\} _{ k\in \{0,1,...,n\}}\)

May be that an illustrative example is useful to understand the topic. In the figure... http://www.123homepage.it/u/i67633995._szw380h285_.jpg.jfif

... we have an increasing function f(x) [the 'blach line'...] that has a single jump discontinuity in $x=.5$. Suppose that n=4, how to construct $\displaystyle f_{4}(x)$?... Mantaining fixed $\displaystyle x_{0}=0,\ x_{2}=.5,\ x_{4}=1$ we are free to schoose in arbitrary way $\displaystyle x_{1}$ and $\displaystyle x_{3}$. At this point we can construct $\displaystyle f_{4}(x)$ connecting the five points with segments [the 'red line'...]. Now we can increase n obtaining $\displaystyle f_{5}(x)$ inserting a new x mantaining fixed the points x=0, x=.5 and x=1 and so one. If the maximum of intevals between the $x_{k}$ tends to 0 if n tends to infinity, then the sequence of function $\displaystyle f_{n}(x)$ pointweis tends to f(x) if n tends to infinity...

Kind regards

$\chi$ $\sigma$