Let f be a function. I am wondering if there exists a function which is not continuous but given by only one functional directive.

i.e. not:

f(x) := $ \begin{cases}1&\text{ for x > 0}\\0&\text{else}\end{cases}$

but something like

f(x) := $e^x$ $\forall x\in \mathbb{R}$.

I think such a function should exists as if it didn't the whole epsilon delta definition of continuity would be superfluous. I am however unable to come up with such a function.

Edit: By functional directive i mean one way to express a function on its domain. In my example the second one would be considered as having only one functional directive. The first one however is defined by two different functional directives for x>0 and x $\leq 0$ seperately.

Edit2: As someone has pointed out this obviously depends on the set of notation techniques that we constrain ourselves to.I think everything that doesn't use multiple functional directives in its own definition should be allowed. The sgn function for example is defined for x > 0, x <0 and x = 0 respectively by three functional directives. It should therefore not be allowed. The function should be elementary by some intuitive notion.

  • $\begingroup$ Wht d u mean by functional directive? $\endgroup$ – mrs Sep 30 '16 at 11:27
  • $\begingroup$ Well, fact is that the thing you say depends on how many symbols we decide to have (which is meaningless, mathematically). For instance, this is a notation every mathematician should be able to understand $$f(x)=\max(\operatorname{sgn} x, 0)$$ which tautologically turns out to be $1$ when $x>0$ and $0$ when $x\le 0$. $\endgroup$ – user228113 Sep 30 '16 at 11:38
  • $\begingroup$ Can a functional directive have limits? Is something like $$f(x)=\lim_{n\to\infty}\sin^{2n}x$$ acceptable? $\endgroup$ – bof Sep 30 '16 at 12:53
  • $\begingroup$ I don't think "one functional directive" is well defined. How do you evaluate $e^{x}$ (as in your example)? Practically speaking you are evaluating a Taylor series ($\infty$-many $+,-,\times,\div$ operations). $\endgroup$ – TravisJ Sep 30 '16 at 13:02
  • $\begingroup$ I don't think the $\epsilon,\delta$ definition of continuity is superfluous if every "nicely expressed" function is continuous. The fact is that, in the space of all continuous functions, only very few of them are "nicely expressible" (i.e. something like a polynomial of any degree, a polynomial composed of trig functions, etc). $\endgroup$ – TravisJ Sep 30 '16 at 13:06

My first suggestion was $$ f(x) = \lfloor x \rfloor \quad (x \in\mathbb{R}) $$

but this seems not to match your expectations.

Alas your definition

By functional directive I mean one way to express a function on its domain.

unfortunately is not a rigorous mathematical definition:

I guess that you mean function definition by express a function on its domain. The one way is too vague for me.

You seem to aim towards defining a continuous function by some base set of continuous functions, combined by operations that keep continuity. Such the result is a continuous function, no doubt.

  • $\begingroup$ see my second edit as to why i do not think this is a valid answer. $\endgroup$ – Zarathustra Sep 30 '16 at 12:32

The following is a concatenation of infinitely many elementary operations that satisfy the conditions asked for:

$f:\mathbb{R}^+ \mapsto \mathbb{R} \cup \{\infty\}$

f(x) = ${{‎‎\sum}}_{n=0}^{\infty‎} x ^n$

The function is not continuous at x = 1 as it converges for all x < 1 but diverges for x $\geq$ 1.

  • 1
    $\begingroup$ Which is $$f(x) = \begin{cases}\frac{1}{1-x} & x < 1 \\ \infty & \text{else}\end{cases}$$ and looks similar to your not-wanted example. $\endgroup$ – mvw Sep 30 '16 at 14:06
  • $\begingroup$ That is indeed true, but the point is that we are able to write it in a different form not using the curly brackets. I may not have made myself clear enough and should have given more thought to the articulation of the question. I do however greatly appreciate your answers. $\endgroup$ – Zarathustra Sep 30 '16 at 14:18

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