Function is Difference of Two Increasing Sequences of Continuous Functions

October 18, 2022 Post a Comment

A sequence of continuous functions converging to a discontinuous function

Solution 1

Take $$f:[0,1]\rightarrow \mathbb{R}\\ f(x)=\left\{\begin{array}{rl} 0 & x\neq 1 \\ 1 & x=1\\ \end{array}\right. $$ And $$f_n=x^n$$ With scaling and piecewise definitions you can to this one for any countable set of $c_1,\dots ,c_n$ In general our function will look like $$f(x)=\left\{ \begin{array}{rl} 0 & x \neq c_i \forall i\\ 1 & \text{else} \\ \end{array}\right.$$ On $[c_i,c_{i+1}]$ we gonna have something like $$f_{ni}(x)=\left(\frac{c_2-x}{c_2-c_1}\right)^n +\left(\frac{x-c_1}{c_2-c_1}\right)^n$$

And all together we will have (with $I=[a,b]$) $$f_n(x)=\left\{ \begin{array}{rl} 0 & x\in [a,c_0)\\ f_{ni} & x \in [c_i,c_{i+1})\\ 0& x\in [c_n,b] \\ \end{array}\right. $$

Edit for a given function the idea is the following, as you only have finite $c_i$ you take with $$\varepsilon=\min_{1\leq i \leq n-1} \{d(c_i,c_{i+1})\}$$ which is the shortest distance between two points of incontinuousity. Edit we don't need Stone Weierstraß at all sry.
$[c_i+\frac{\varepsilon}{2n},c_{i+1}-\frac{\varepsilon}{2n}]$ we just take $f$ on the intervalls (the uniform convergence is trivial). So we only need to chose a secquence of function on $[c_i-\frac{\varepsilon}{2n},c_i+\frac{\varepsilon}{2n}]$. We will call them $s_{ni}$ (like spline).
We chose $$s_{ni}(x)= \left\{ \begin{array}{rl} f(c_i) + \frac{f\left(c_i-\frac{\varepsilon}{2n}\right)-f(c_i)}{\frac{\varepsilon}{2n}} \cdot (x-c_i) & x-c_i \leq 0\\ f(c_i)+\frac{f\left(c_i+\frac{\varepsilon}{2n}\right)-f(c_i)}{\frac{\varepsilon}{2n}}\cdot (x-c_i) & x-c_i >0 \end{array}\right.$$ Ok that one looks really complicated but all i am saying we make a line from the left end to the point we want to have $f(c_i)$ and another one to get a continuous function in all the intervall.

Solution 2

The general idea is best explained when there is only one discontinuity point:

Suppose $f$ is continuous for all $x\ne b$. Select two points $a<b$ and $c>b$. Define the function $g$ by setting $g(x) =f(x)$ for $x\notin (a,c)\setminus\{b\}$, and on $ (a,c)\setminus\{b\}$, take the graph of $g$ to be piecewise linear. Do this is such a way that $g$ is continuous.

You should draw the picture here. You're just replacing the "discontinuous part" of the graph of $f$ with straight line segments; thus producing a continuous function that agrees with $f$ except on an interval of small length. For the pointwise convergence of the sequence to come, it is important to have $g(b)=f(b)$, here.

By selecting sequences $(a_n)$ and $(b_n)$ with $a_n\nearrow b$ and $c_n\searrow b$, and defining continuous functions $g_n$ as above, one obtains a sequence of continuous functions that converge pointwise to $f$.

If $f$ has finitely many points of discontinuity, $x_1$, $\ldots\,$, $x_k$, do the same thing:

For each $n$, select intervals $O_1$, $\ldots\,$, $O_k$, such that:

$\ \ \ $1) The $O_i$ are pairwise disjoint.

$\ \ \ $2) The sum of the lengths of the $O_i$ is at most $1/n$.

$\ \ \ $3) For each $k$, $x_k$ is the midpoint of $O_k$.

Now define your $f_n$ appropriately.

Comments

Let $f:I\to \mathbb R$ be a function which is continuous in every points of the interval $I$ except of a finite number of discontinuities $c_1,...,c_n$. I would like to find a sequence of continuous functions $f_n:I\to \mathbb R$ such that $\lim f_n=f$ pointwise.

This question seems very difficult, maybe because this one is very general, I'm really stuck here, any help is welcome.

Thanks a lot
- For a problem like this, I think it's easiest to start with an example of such an $f$ (and keep it simple, say one discontinuity) and then come up with a series of continuous functions $f_n$ that make better and better approximations of $f$. Once you do this for an example, you have a better idea how to do it for any such $f$.
- @ChrisPhan yes, I've already tried with one discontinuity, without any success
- Do you have any more information on the discountinuities?
- @George no just the information I've already mentioned above.
- The most annoying case is if you have a vertical asymptote $\{x=x_0\}$. Draw a picture. Use broken lines on $[x_0-1/n,x_0+1/n]$. Make it pass through $(x_0,f(x_0))$. Note the method of broken lines works in general.
- For each $n$, take closed intervals that are pairwise disjoint of small length, such that each interval contains a discontinuity point of $f$ in its interior (and the interval contains no other discontinuity point of $f$). Define $f_n$ to be $f$ off the union of these intervals. On the union of the intervals, define $f_n$ to be piecewise linear appropriately (in particular, define $f_n(y)=f(y)$ for a discontinuity point $y$ of $f$).
- @DavidMitra can you make an answer with a little bit more details, thank you for your comment
- @DavidMitra What about the case when there're more discontinuities, say countable (for example, $\mathbb Q$)?
- @julien: David Mitra's approach takes care of this, but for what it's worth you can initially consider the arctangent of the function, only deal with finite discontinuities, and then take the tangent of the resulting continuous functions.
- @Frank Science: Such a sequence of continuous functions can be found for any ${\cal F}_{\sigma}$ meager set. (${\cal F}_{\sigma}$ means it can be written as a countable union of closed sets and meager means it can be written as a countable union of nowhere dense sets.) See Theorem 3' in my post on Continuity Sets. Note: An ${\cal F}_{\sigma}$ meager set can have cardinality continuum in each interval, or even have full Lebesgue measure in each interval, or even have a Hausdorff dimension zero complement.
- @FrankScience For countably many discontinuities, the result is still true. But, I don't think my approach for the finite case will work here. See this though.
The problem is $f$ is a general one, I'm not looking for an specific example, thank you
ah you didn't got the idea
no, sorry. My problem is given a $f$ (could be any $f$ with the specifications above) I would like to find a $f_n$ such that $f_n$ goes to $f$. Did you follow me?
but thank you anyway for trying to help me.
yeah i did follow you i edited it, thought it would be homework so I didn't want to do everything
no it's not a homework, it's to prepare to an exam tomorrow.
is it fine now ? its nearly the same as before
yes, your solution is really fine, and really helpful, I'm studying right now your example, thank you. However the only problem is not what I'm asking.

Recents

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Source: https://9to5science.com/a-sequence-of-continuous-functions-converging-to-a-discontinuous-function

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