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You are given a function $f: \mathbb{R}\rightarrow \mathbb{R}$. Every derivative $\frac{d^n}{dx^n}(f(x)), \,n >0$ of the function is continuous.

Is there a function $g: \mathbb{R}\rightarrow \mathbb{R}$, for which every derivative $\frac{d^n}{dx^n}(g(x)), \,n >0$ is also continuous, such that: $$\forall x\in[a,b]: \, g(x) = f(x)\land \, \exists x \notin [a,b]: f(x) \neq g(x),\, a \neq b$$

Thanks!

Edit:

I asked the question because I intuitively wondered if there would be functions, which could behave like "the same" in a given interval, but then behave differently so that they start diverging or at least stopped being the same the anymore. The answer to this question gave me a bigger understanding of real analysis.

If you would like to know which made me think about such a problem: Although this is a vague formulation, generally, this question asks if two (completely) different things can develop (themselves) to be exactly equal in at least one part of there whole existence...

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    $\begingroup$ FYI, this can't happen with complex-differentiable $f: \mathbb{C} \to \mathbb{C}$. Such functions are immediately infinitely often differentiable, so it indeed matches your setting. If two complex-differentiable functions agree on an open set, they already agree everywhere on their domain. $\endgroup$ – ComFreek Apr 13 at 13:58
  • $\begingroup$ How to proof that? $\endgroup$ – TVSuchty Apr 13 at 17:28
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    $\begingroup$ The proof of the so-called "identity theorem" is usually based on several previous lemmas. I think this comment section is too small for that, but I suggest having a look at identity theorem @ Wikipedia or in any introductory complex analysis book. $\endgroup$ – ComFreek Apr 15 at 6:35
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Define the real functions $f$ and $g$ thus: $$ f(x) = \begin{cases} \exp\Big(-\frac{1}{(x - 1)^2}\Big)\ &\text{if } x > 1 \\ 0\ &\text{if } x \in [-1, 1] \\ \exp\Big(-\frac{1}{(x + 1)^2}\Big)\ &\text{if } x < -1 \end{cases} $$ and $$ g(x) = 0\quad \text{for all } x \in \mathbb{R} $$

  • By construction, $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.

  • Obviously $g$ is continuously differentiable infinitely many times as it is a constant function.

  • You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by computing appropriate right and left hand limits of $\frac{d^nf(x)}{dx^n}$ ($n \in \mathbb{Z}_+$) inductively. Let us compute for example $\lim\limits_{x \to 1^+}\frac{df(x)}{dx}$: for $x > 1$, \begin{align*} 0 < \frac{df(x)}{dx} &= \frac{2(x - 1)^{-3}}{\exp\big(\frac{1}{(x - 1)^2}\big)} \\ &= \frac{2y^3}{\exp\big(y^2\big)} \quad\text{ where }y = \frac{1}{x - 1} > 0 \\ &= \frac{2}{\frac{1}{y^3}\sum\limits_{k = 0}^\infty \frac{y^{2k}}{k!}} \quad\text{ divide by }y^3\text{ and use }e^z = \sum\limits_{k = 0}^\infty\frac{z^k}{k!} \\ &= \frac{2}{\frac{1}{y^3}\frac{y^0}{0!} + \frac{1}{y^3}\frac{y^2}{1!} + \frac{1}{y^3}\frac{y^4}{2!} + \sum\limits_{k = 3}^\infty \frac{y^{2k - 3}}{k!}} \\ &< \frac{2}{\frac{1}{y^3} + \frac{1}{y} + \frac{1}{2}y} \end{align*} As $x \to 1^+$, we get $y = \frac{1}{x - 1} \to \infty$ and the right hand fraction $\frac{2}{\frac{1}{y^3} + \frac{1}{y} + \frac{1}{2}y} \to 0$. So, as $x \to 1^+$, $\frac{df(x)}{dx} \to 0$ also by the Squeeze Theorem.

That was a long calculation but take my word: it can be repeated inductively to show that $\lim\limits_{x \to 1+}\frac{d^nf(x)}{dx^n} = 0$ for all $n \in \mathbb{Z}_+!$ At all other points i.e. on $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.

Bonus Fact:

Both $\frac{d^n f(x)}{dx^n}$ and $\frac{d^n g(x)}{dx^n}$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!

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    $\begingroup$ It basically says under certain conditions, $\lim\limits_{x \to a}(f(x) / g(x)) = \lim\limits_{x \to a}(\frac{d f(x)}{dx} / \frac{d g(x)}{dx})$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule $\endgroup$ – 0XLR Apr 12 at 22:49
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    $\begingroup$ I am stunned. Do you know of more complex solutions? $\endgroup$ – TVSuchty Apr 12 at 22:52
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    $\begingroup$ @TVSuchty what class shows you logic symbols before explaining L'Hôpitals rule? $\endgroup$ – uhhhhidk Apr 13 at 0:45
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    $\begingroup$ We can obtain $f^{(n)}(1)=0$ for all $n$ without l'Hopital's Rule by induction and some elementary considerations........+1 $\endgroup$ – DanielWainfleet Apr 13 at 6:41
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    $\begingroup$ @DanielWainfleet Ah yes, I should probably add an edit with an elementary proof since the OP admitted he did not cover L'Hôpital. $\endgroup$ – 0XLR Apr 13 at 6:59
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Sure. In fact, there is a whole class of functions which not only exist, but are specifically made to do something that effectively implies exactly what you are looking for: they are called bump functions, and are defined as smooth (differentiable everywhere, arbitrarily many times) functions that have compact support, meaning (almost) that they are zero everywhere except on a compact set, which for the real numbers as domain basically means a closed, bounded (i.e. contained within an interval) subset thereof, such as a closed interval. This compact set where they are nonzero is called the "support". The trick is to exploit the "everywhere else zero"-ness, as that gives you what you're after.

Namely, any two different (i.e. not equal) bump functions with the same supporting interval $[a, b]$, will be smooth, zero on any interval outside this interval, and yet different, because they differ on such interval. More generally, given any two different bump functions, period, you just have to find an interval outside both of their support sets, which is always possible because they are both bounded.

A simple example of such a bump function is

$$\mathrm{bump}: \mathbb{R} \rightarrow \mathbb{R},\ \ \ \ \mathrm{bump}(x) := \begin{cases} e^{-\frac{1}{1 - x^2}},\ \mbox{when $x$ is in $(-1, 1)$}\\ 0,\ \mbox{otherwise} \end{cases}$$

Then consider just $\mathrm{bump}(x)$ and a nonzero multiple thereof, say, $\mathrm{bump2}(x) := 2 \cdot \mathrm{bump}(x)$. We now have $\mathrm{bump}(x) = \mathrm{bump2}(x)$ when, say, $x \in [10, 11]$, since they are both zero there. Yet, they are ostensibly not equal when $x$ is in $(-1, 1)$.

ADD: It appears to have been asked as to how one can do this without explicitly constructing the bump function. The above is just to show (not completely thoroughly) that bump functions exist. Indeed, we can do so as well. Let now $\mathrm{bump}(x)$ be a general bump function. Let its support set be $\mathrm{supp}[\mathrm{bump}]$. That is,

$$\mathrm{supp}[\mathrm{bump}] := \mathrm{cl}\left(\{ x \in \mathbb{R} : \mathrm{bump}(x) \ne 0 \}\right)$$

(n.b. "cl" means to take the closure; basically this includes all "endpoints" of regions in which it is nonzero, even if it is zero at those endpoints - e.g. the support of the just-given-explicitly bump function is $[-1, 1]$, not $(-1, 1)$. This is a bit of technicality that was wrapped earlier when I said "almost" in "meaning (almost)" above.)

Since the support set is bounded and closed, it has a maximum and minimum (largest and smallest element): assign $M := \mathrm{max}\ \mathrm{supp}[\mathrm{bump}]$. Now consider the interval $\mathrm{ext\ ival} := [M+1, M+2]$. If $x \in \mathrm{ext\ ival}$, then it is clearly not in the support set, but rather to the right of it. Thus $\mathrm{bump}(x) = 0$ there. Now set $\mathrm{bump2}(x) := 2 \cdot \mathrm{bump}(x)$ as before (if you want even more generality, just replace $2$ with an arbitrary vertical rescaling coefficient $a$ that is not $0$ or $1$). Congrats, you now have two bump functions that are unequal but equal on the external interval $\mathrm{ext\ ival}$.

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    $\begingroup$ I see what you did there! Is there a case, where you can write an example without defining the function bit by bit, maybe just as term? $\endgroup$ – TVSuchty Apr 13 at 9:08
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    $\begingroup$ @TVSuchty : Actually, yes. See what I just added. I believe this is what you're asking about. $\endgroup$ – The_Sympathizer Apr 13 at 14:21
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    $\begingroup$ @TVSuchty : "I see what you did there!" yeah I just made the little bumpzzle perk up :) #mehhr. $\endgroup$ – The_Sympathizer Apr 13 at 14:45
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    $\begingroup$ @TVSuchty piecewise defined functions are not special and you shouldn't see functions "just as term" as not being piecewise either. $\endgroup$ – Tony Apr 21 at 16:42

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