0

Could non-smooth time-limited functions been Analytical? Please read the scenarios first

I was reading about analytic functions definitions on Wiki and looks like some of its properties where Differentiabilty is involved depends on where is the function defined on $\mathbb{R}$ or $\mathbb{C}$, and also reading about Bump functions $\in C_c^\infty$ on Wiki and Non-analytic smooth functions here, it is said that "One can easily prove that any analytic function of a real argument is smooth. The converse is not true" and also if a function is smooth and compact-supported, then it cannot be an analytic function.

So here is where I get kind of lost: if I think in $f(t)$ as a continuous time-limited function (so is compact-supported with boundaries $\partial t = \{t_0,\,t_F\}$, and also, so is continuous and compact-supported, it is also a bounded function)... since is compacted supported, if I let it restricted to be a "smooth function" I will already know it couldn't be analytical, ¿but what will happen if is not smooth?

So, to understand what could happen, given differentiability could be an domain-related issue, I want to know what could happen in each of the following scenarios:

  1. Let $f(t)$ been a continuous time-limited function such as $f(t)\in\mathbb{R}$ that is not smooth (even, it could being neither a differentiable function), fulfilling also that $f(t_0)=f(t_F)=0$... Could this $f(t)$ been an analytical function?
  2. Let $f(t)$ been a continuous time-limited function such as $f(t)\in\mathbb{R}$ that is not smooth (even, it could being neither a differentiable function), where it can have any of their values at the edges of the domain $f(t_0)\neq 0$ and/or $f(t_F)\neq 0$ (not necessarily both equals, so $f(t_0) \neq f(t_F)$ in general)... Could this $f(t)$ been an analytical function?
  3. Let $f(t)$ been a continuous time-limited function such as $f(t)\in\mathbb{C}$ that is not smooth (even, it could being neither a differentiable function), fulfilling also that $f(t_0)=f(t_F)=0$... Could this $f(t)$ been an analytical function?
  4. Let $f(t)$ been a continuous time-limited function such as $f(t)\in\mathbb{C}$ that is not smooth (even, it could being neither a differentiable function), where it can have any of their values at the edges of the domain $f(t_0)\neq 0$ and/or $f(t_F)\neq 0$ (not necessarily both equals, so $f(t_0) \neq f(t_F)$ in general)... Could this $f(t)$ been an analytical function?

PS: the "edges-part" is because if they are non-zero, it cannot be smooth and compacted-supported under no-other conditions, and also since they made a discontinuity in the "transition", it creates some issues in its derivatives (where some delta functions arise), and also in their Fourier transforms, and I am interested if this issue has something to said within the topic of this question.

As example, $$f(t) = \begin{cases} \cos^4\left(\frac{t\pi}{2}\right),\,|t|\leq 1 \\ 0,\,|t|>1 \end{cases}$$ is a real valued function with compact-support that is not smooth (because is not true that $\lim\limits_{t \to \partial t} \frac{df^n(t)}{dt^n} = 0\,\forall n \geq 0$), but it is analytical "within" its support (i.e., within the open interval $(-1,\,1)$), and its transition to the boundaries is "soft" (meaning here, that the $f(t)$ and $f'(t)$ are continuous and equal to zero at $t = \{-1,\,1\}$ with $\lim\limits_{t \to \partial t^-} f(t) = \lim\limits_{t \to \partial t^+} f(t)$ and $\lim\limits_{t \to \partial t^-} f'(t) = \lim\limits_{t \to \partial t^+} f'(t)$), here the plot of $f(t)$, it looks actually really "good behaved", so I want to know if these kind of functions could also be analytic - or if not.

Joako
  • 1,380
  • 1
    If a function is analytic it is smooth. Therefore 1,2,3,4 the answer is no. Once you say the function is not smooth that is it, it cannot be analytic. – Snaw Dec 05 '21 at 02:10
  • @Snaw thanks for answering. Maybe you are right, but from Wikipedia quote I cite at the beginning, it is said that "One can easily prove that any analytic function of a real argument is smooth", so, the argument part is an unnecessary detail? or your answer is valid only for real argument functions? – Joako Dec 05 '21 at 02:13
  • 1
    By a real argument they mean a function which takes real numbers as inputs, not necessarily that the domain is the whole of $\mathbb{R}$. – Snaw Dec 05 '21 at 02:21
  • @Snaw since the domain could be different from the whole $\mathbb{R}$, and time-limited cannot being analytic: They are going to be non-analytic just on the boundaries points $t_0$ and $t_F$? or non-analytical on the whole disjoint intervals $(-\infty,,t_0] \cup [t_F,,\infty)$?? – Joako Dec 05 '21 at 02:30
  • 2
    For your $f(t)$ function example, it is not smooth on the whole of $\mathbb{R}$ and indeed it is also not analytic there. If you look at $(-1,1)$ then it is analytic there (and therefore also smooth). Analytic in an open interval means that it is analytic for every point of the open interval which means that for every point of the open interval there is a neighbourhood in which the function equals its Taylor series around that point. In that sense being analytic is a pointwise property. Does this clarify things? – Snaw Dec 05 '21 at 02:39
  • Yes, but I was believing (wrongly I guess), that the domain where is zero it was also analytical (so just in the edges of the domain the function was failing to be analytical), since the zero function can be described by a Taylor function with zero coefficients.., but maybe it was a dumb assumption. – Joako Dec 05 '21 at 03:11
  • 1
    Yes, you are correct. It would be analytic for every x<-1 and for every x>1 as well, since there for every $x$ there is a neighbourhood around x in which $f(x)$ is the constant zero function which naturally equals its Taylor expansion. – Snaw Dec 05 '21 at 05:19
  • 1
    FYI - that "of a real argument" is indeed an unnecessary detail. "Analytic" implies a function on either $\Bbb R^n$ or $\Bbb C^n$ for any $n$ is infinitely differentiable. The classic example of a function which is smooth (in the sense of infinite differentiable) but not analytic is $$f(x) = \begin{cases}0& x \le 0\e^{-1/x}& x > 0\end{cases}$$which is smooth everywhere and analytic on $(-\infty, 0)$ and $(0,\infty)$, but is not analytic at $0$. $f^(k)(0) = 0$ for all $k$, so the Taylor series at $0$ is just $0$, which does not equal $f(x)$ for any $x > 0$. – Paul Sinclair Dec 05 '21 at 21:52

1 Answers1

0

As it was commented by others on the question, Analytic functions must be smooth, so, if the function isn´t smooth, it can´t be an Analytic function.

I just left the answer explicitly to close the question.

Joako
  • 1,380