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After some investigation it seems fairly obvious to me that the only such function is the zero function, however I haven't been able to prove it. By considering $$\alpha =\sup\{x\in[0,+\infty) :f(x) = 0\},$$ I was able to show that $\alpha$ can only be $1$ or $0$ but I could not weed out those two possibilities. Any hints/solutions welcome.

EDIT 1

Because of the continuity of $f$, we must have $f(\alpha) = 0$. Note that because of the relation given we have $$\int_0^{\sqrt \alpha}2xf'(x^2)\,\mathrm dx = f(\alpha),$$ but because of the relationship given this implies $$\int_0^{\sqrt \alpha}2xf(x)\,\mathrm dx = f(\alpha).$$ If $\alpha$ is strictly between $0$ and $1$, then $\sqrt \alpha > \alpha$, but then splitting the integral we get $$\int_{\alpha}^{\sqrt \alpha}2xf(x)\,\mathrm dx = f(\alpha) = 0.$$ But by our choice of $α$, this integral should be non-zero since our function is positive. Hence $\alpha$ cannot be between $0$ and $1$.

Now suppose it is greater than $1$, then we have $$f(\alpha^2) =\int_0^{\alpha}2xf(x)\,\mathrm dx = 0.$$ Since our function is $0$ on $[0,\alpha]$ (Note that it is increasing), this is again a contradiction because $\alpha^2 > \alpha$. Therefore $\alpha$ is $0$ or $1$.

EDIT 2

I forgot to mention the important condition that $f(0)=0$.

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  • $\begingroup$ Is this a problem you were given, or is it one that you came up with? Should we expect the existence of some neat solution here? $\endgroup$ – Omnomnomnom May 31 '18 at 12:28
  • $\begingroup$ If you have actually proven that, since f is non decreasing (f' >= 0 ) , it's easy to finish: if it's 1, then since f is continuous and non decreasing it can't be it, and the same goes with 0. But how did you end up with these two possibilties? haven't thought about it the slightest though.. $\endgroup$ – mvggz May 31 '18 at 13:03
  • $\begingroup$ @mvggz see my edit. I don’t quite understand your reasoning as to why 0 and 1 can’t be the sup $\endgroup$ – Quantaliinuxite Jun 1 '18 at 13:41
  • $\begingroup$ @Quantaliinuxite Ok , first of all I like your approach :) , my argument relies mainly on the fact that f is non decreasing and continuous , thanks to the relationship given ( f' >= 0 ). What you get from this is : $\forall x \in [0, +\infty[$ , $f(x) \geq f(0) $ So if $\alpha = 0$ , $f(x) = 0 , \forall x \in [0, +\infty[$. If $\alpha = 1$, in particular $\forall x \in [1, +\infty[$ , $f(x) \geq f(1) = f(\alpha) $ and since f is continuous, $f( \alpha + \epsilon) \geq f(\alpha)$. Hence $\alpha$ is not the sup defined as such. $\endgroup$ – mvggz Jun 1 '18 at 14:13
  • $\begingroup$ I don't understand how you get your last equality though, the one with $\alpha^{2}$. I can get: $f(\alpha^2) = \int_{0}^{\alpha} 2xf^{'}(x^2)dx = \int_{0}^{\sqrt\alpha} 2xf^{'}(x^2)dx + \int_{\sqrt\alpha}^{\alpha} 2xf^{'}(x^2)dx = \int_{\sqrt\alpha}^{\alpha} 2xf^{'}(x^2)dx$ But then how do you from here? $\endgroup$ – mvggz Jun 1 '18 at 14:26
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As Alex already noticed, a slightly more general statement holds:

Let $f:[0, \infty) \to [0, \infty)$ be continuous, differentiable on $(0, \infty)$, and $c \ge 1 $.

If $f(0) = 0$ and $f(x) = f'(x^c)$ for all $x > 0$ then $f = 0$.

Proof: $f'(x) = f(x^{1/c}) \ge 0$, so that $f$ is increasing.

This in turn implies that $f'$ is increasing on $(0, \infty)$, so that $f$ is convex.

Step 1: $f(x) = 0$ for $0 \le x \le 1$.

From the convexity and $f(0) = 0$ it follows that $$ f(t) \le t \cdot f(1) \quad \text{ for } 0 \le t \le 1 \, . $$ On the other hand, the mean-value theorem gives $$ f(1) - f(0) = f'(\xi) (1 - 0) $$ for some $\xi \in (0, 1)$, therefore $$ f(1) = f'(\xi) = f(\xi^{1/c}) \le \xi^{1/c} \cdot f(1) \, . $$ $\xi^{1/c}$ is strictly less than one, so that $f(1) \le 0$ follows.

Since $f$ is increasing, $f(x) = 0$ for $0 \le x \le 1$.

Step 2: $f(x) = 0$ for $x \ge 1$.

For $x \ge 1$ $$ f'(x) = f(x^{1/c}) \le f(x) $$ so that we can use a standard (Grönwall's inequality type) argument: $h(x) = e^{-x} f(x)$ satisfies $$ h'(x) = e^{-x} (f'(x) - f(x)) \le 0 $$ so that $h$ is decreasing on $[1, \infty)$: $$ e^{-x} f(x) \le e^{-1} f(1) = 0 \\ \implies f(x) \le 0 \implies f(x) = 0 \, . $$

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  • $\begingroup$ It's interesting to note that the case $c=1$ is just the easy to solve differential equation $f(x)=f'(x)$, whose general solution is just $f(x)=Ce^x$, $C\in \mathbb R$. The initial condition $f(0)=0$ corresponds to the particular solution $f(x)\equiv 0$. So it is not contrary to intuition that it also be the case for $c>1$ (since for $x_0\in (0,1)$ the value of $f(x_0)$ is determined by the value of the derivative at $x_0^2<x_0$, that is, from the values of $f$ in a neighborhood where 'it has already been defined'). $\endgroup$ – Alejandro Nasif Salum Jun 5 '18 at 7:27
  • $\begingroup$ @AlejandroNasifSalum: Actually “Step 1” works for all $c > 0$, so one can conclude that $f=0$ on $[0, 1]$ even if $0 < c < 1$. $\endgroup$ – Martin R Jun 5 '18 at 9:25
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    $\begingroup$ This answer is by far the simplest, and most to the point. I wasn’t aware of the $e^{-x}$ trick, so that was useful also. $\endgroup$ – Quantaliinuxite Jun 8 '18 at 13:13
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EDIT The following post was made before the condition $f(0)=0$ was stated, which leaves my critique and my counterexample inapplicable. I leave it here since I find it of interest in itself, and because if the proposed conjectures $f(x)=f(1)\cdot f_1(x)$ and $\forall x\ge 0,\;\textrm{sgn} f(x) = \textrm{sgn} f(1)$ (see below) were true, it would imply that the condition $f(0)=0$ is necessary for the conclusion $f(x)\equiv0$ to hold, and this zero function would just be a particular solution of the 'functional-differential' equation $f(x)=f'(x^2)$.


Your reasoning has a gap at the very beginning: the supremum of the set $\{x\in[0,\infty)\colon f(x)=0\}$ exists if it is both bounded above and nonempty. I think it would not be difficult to see that if it is nonempty and $f(x)\not\equiv 0$ then it will be bounded above, but I don't see why it should be nonempty anyway (unless we add the condition that $f$ be surjective).

By the way, I did some numerical approximation taking $f(1)=1$ as 'initial' condition, and I ended up with this $f$:

enter image description here

Here are some values.

$$\begin{array}\\x & y\\ 0.0 &0.2887337\\ 0.5 &0.5656723\\ 1.0 &1.0000000\\ 1.5 &1.5602165\\ 2.0 &2.2340116\\ 2.5 &3.0138627\\ 3.0 &3.8944997\\ 3.5 &4.8719327\\ 4.0 &5.9429892\\ 4.5 &7.1050584\\ 5.0 &8.3559366\\ \end{array}$$

The problem seems to be well conditioned and the behavior of the iterative procedure looked stable. Moreover, other functions $f$, for different initial values at $x=1$ seem to be multiples of the one given above (say $f_1$), in fact, of the form $$f(x)=f(1)\cdot f_1(x).$$ Also $f_1$ is likely positive, so the last equation and this condition would imply $$\forall x\ge 0,\;\textrm{sgn} f(x) = \textrm{sgn} f(1).$$

On the other hand, I still don't see a reasonable closed form expression for such a function.

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  • $\begingroup$ If the best guess is correct, the reasonable closed form is $f(x)=0.$ $\endgroup$ – Somos Jun 3 '18 at 4:18
  • $\begingroup$ Indeed I am reasoning by contradiction since I am trying to show that set is unbounded. The numerical approximation is interesting but I tend to be wary of them in such cases where existence hasn’t been established prior. $\endgroup$ – Quantaliinuxite Jun 3 '18 at 6:35
  • $\begingroup$ Note also that it is non empty since 0 belongs to the set from our initial conditions. Also it $f$ is non zero it must surjective since the function is increasing (as are all the subsequent derivatives) $\endgroup$ – Quantaliinuxite Jun 3 '18 at 6:45
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    $\begingroup$ @AlexFrancisco That's my mistake, I'll add it $\endgroup$ – Quantaliinuxite Jun 3 '18 at 7:32
  • $\begingroup$ See my new revised answer for the Puiseux series for your function on $0\le x\le 1.$ $\endgroup$ – Somos Jun 6 '18 at 3:07
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$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$A generalized proposition will be proved.

Proposition: For any given $c > 1$, if $f: [0, +∞) → [0, +∞)$ is continuous on $[0, +∞)$, differentiable on $(0, +∞)$, and $f'(x^c) = f(x)\ (x > 0)$, $f(0) = 0$, then $f = 0$.

Step 1: For any $0 \leqslant a < b$, $n \geqslant 0$,\begin{align*} f(b) - f(a) &= \sum_{k = 1}^n (-1)^{k - 1} \left( \prod_{j = 1}^k \frac{c - 1}{c^j - 1} \right) (b^{\frac{c^{k + 1} - c}{c^{k + 1} - c^k}} f(b^{\frac{1}{c^k}}) - a^{\frac{c^{k + 1} - c}{c^{k + 1} - c^k}} f(a^{\frac{1}{c^k}}))\\ &\peq + (-1)^n \left( \prod_{j = 1}^n \frac{c - 1}{c^j - 1} \right) \int_{a^{\frac{1}{c^n}}}^{b^{\frac{1}{c^n}}} x^{\frac{c^{n + 1} - c}{c - 1}} f'(x) \,\d x. \tag{1} \end{align*}

Proof: To prove by induction on $n$, the base case $n = 0$ is true because$$ f(b) - f(a) = \int_a^b f'(x) \,\d x. $$

Assume that it holds for $n - 1$. For $n$, note that $f'(x^c) = f(x)$. By integration by parts,\begin{align*} &\peq \int_{a^{\frac{1}{c^{n - 1}}}}^{b^{\frac{1}{c^{n - 1}}}} x^{\frac{c^n - c}{c - 1}} f'(x) \,\d x = \int_{a^{\frac{1}{c^n}}}^{b^{\frac{1}{c^n}}} t^{\frac{c^{n + 1} - c^2}{c - 1}} f'(t^c) · ct^{c - 1} \,\d t = c \int_{a^{\frac{1}{c^n}}}^{b^{\frac{1}{c^n}}} t^{\frac{c^{n + 1} - c}{c - 1} - 1} f(t) \,\d t\\ &= \left. \frac{c - 1}{c^n - 1} t^{\frac{c^{n + 1} - c}{c - 1}} f(t) \right|_{a^{\frac{1}{c^n}}}^{b^{\frac{1}{c^n}}} - \int_{a^{\frac{1}{c^n}}}^{b^{\frac{1}{c^n}}} t^{\frac{c^{n + 1} - c}{c - 1}} f'(t) \,\d t\\ &= \frac{c - 1}{c^n - 1} (b^{\frac{c^{n + 1} - c}{c^{n + 1} - c^n}} f(b^{\frac{1}{c^n}}) - a^{\frac{c^{n + 1} - c}{c^{n + 1} - c^n}} f(a^{\frac{1}{c^n}})) - \int_{a^{\frac{1}{c^n}}}^{b^{\frac{1}{c^n}}} x^{\frac{c^{n + 1} - c}{c - 1}} f'(x) \,\d x. \end{align*} Combining with the induction hypothesis, it holds for $n$. End of induction.

Step 2: $f(1) = 0$.

Proof: For any $n \geqslant 0$, set $a = 0$ and $b = 1$ in (1) to get\begin{align*} f(1) &= \sum_{k = 1}^n (-1)^{k - 1} \left( \prod_{j = 1}^k \frac{c - 1}{c^j - 1} \right) f(1) + (-1)^n \left( \prod_{j = 1}^n \frac{c - 1}{c^j - 1} \right) \int_0^1 x^{\frac{c^{n + 1} - c}{c - 1}} f'(x) \,\d x\\ &= \sum_{k = 1}^n (-1)^{k - 1} \left( \prod_{j = 1}^k \frac{c - 1}{c^j - 1} \right) f(1) + (-1)^n \left( \prod_{j = 1}^n \frac{c - 1}{c^j - 1} \right) \int_0^1 t^{\frac{c^{n + 2} - c^2}{c - 1}} f'(t^c) · ct^{c - 1} \,\d t\\ &= \sum_{k = 1}^n (-1)^{k - 1} \left( \prod_{j = 1}^k \frac{c - 1}{c^j - 1} \right) f(1) + (-1)^n c \left( \prod_{j = 1}^n \frac{c - 1}{c^j - 1} \right) \int_0^1 t^{\frac{c^{n + 2} - c}{c - 1} - 1} f(t) \,\d t. \end{align*} Denote $c_n = \prod\limits_{j = 1}^n \frac{c - 1}{c^j - 1}\ (n \geqslant 0)$, then$$ f(1) = f(1) \sum_{k = 1}^n (-1)^{k - 1} c_k + (-1)^n c c_n \int_0^1 t^{\frac{c^{n + 2} - c}{c - 1} - 1} f(t) \,\d t. \tag{2} $$ Note that $\{c_n\}$ is strictly decreasing and $c_n → 0\ (n → ∞)$, then $\sum\limits_{k = 0}^∞ (-1)^{k - 1} c_k$ converges and $\sum\limits_{k = 0}^∞ (-1)^{k - 1} c_k < c_0 = 1$. Also, suppose that $|f(x)| \leqslant M$ for $0 \leqslant x \leqslant 1$, then$$ \left| (-1)^n c c_n \int_0^1 t^{\frac{c^{n + 2} - c}{c - 1} - 1} f(t) \,\d t \right| \leqslant c c_n M, $$ which implies$$ \lim_{n → ∞} (-1)^n c c_n \int_0^1 t^{\frac{c^{n + 2} - c}{c - 1} - 1} f(t) \,\d t = 0. $$ Thus, making $n → ∞$ in (2) to get$$ f(1) = f(1) \sum_{k = 1}^∞ (-1)^{k - 1} c_k. $$ Since $\sum\limits_{k = 1}^∞ (-1)^{k - 1} c_k < 1$, then $f(1) = 0$.

Step 3: $f = 0$.

Proof: Note that $f'(x) = f(x^{\frac{1}{c}}) \geqslant 0$, thus $f$ is increasing. Since $f(0) = f(1) = 0$, then $f(x) = 0$ for $0 \leqslant x \leqslant 1$. For $b > 1$, taking $a = 1$ and $n = 1$ in (1) to get$$ f(b) = b f(b^{\frac{1}{c}}) - \int_1^{b^{\frac{1}{c}}} x^c f'(x) \,\d x \leqslant b f(b^{\frac{1}{c}}). $$ Note that $b > 1 \Rightarrow b^{\frac{1}{c}} > 1$. By induction,$$ f(b) \leqslant b^{\frac{c^n - 1}{c^n - c^{n - 1}}} f(b^{\frac{1}{c^n}}). \quad \forall n \geqslant 1 \tag{3} $$ Since $\dfrac{c^n - 1}{c^n - c^{n - 1}} → \dfrac{c}{c - 1}\ (n → ∞)$ and $b^{\frac{1}{c^n}} → 1\ (n → ∞)$, making $n → ∞$ in (3) to get$$ f(b) \leqslant \lim_{n → ∞} b^{\frac{c^n - 1}{c^n - c^{n - 1}}} · \lim_{n → ∞} f(b^{\frac{1}{c^n}}) = b^{\frac{c}{c - 1}} f(1) = 0, $$ which implies $f(b) = 0$. Therefore, $f = 0$.

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  • $\begingroup$ Once you have $f(1)=0$ you can also use that $f'(x) = f(x^{\frac{1}{c}}) \le f(x)$ for $x \ge 1$, and a standard argument (considering the derivative of $e^{-x}f(x)$) shows that $f = 0$. $\endgroup$ – Martin R Jun 3 '18 at 11:57
  • $\begingroup$ @MartinR That's smart! I was only thinking about keeping using step 1 till the end and didn't jump out of it. $\endgroup$ – Saad Jun 3 '18 at 12:00
  • $\begingroup$ This looks very good, though the argument is very intricate, especially the first part by induction. Can I ask what clues made you think of it? $\endgroup$ – Quantaliinuxite Jun 3 '18 at 16:20
  • $\begingroup$ @Quantaliinuxite Actually step 1 is only useful for the case $a=0$, $b=1$ for proof ensuing, but to incorporate the case $n=1$ in the step 3, I proved the general identity for convenience, $\endgroup$ – Saad Jun 4 '18 at 2:17
  • $\begingroup$ @Quantaliinuxite The idea comes from the fact that $x=0,1$ are the fixed points of the function $x^c-x$. $\endgroup$ – Saad Jun 4 '18 at 2:19
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My previous work was based on a false premise that $\,f(x)\,$ has a power series expansion at $\,x=0.\,$ The key insight is the correct ansatz. Since the differential equation is homogeneous, if $\,f(0) \ne 0, \,$ then WLOG assume $\, f(0)=1. \,$ Also assume that $\, f(x) \,$ has a Puiseux series expansion. Start with $\, f(x) = \sum_{n=0}^\infty a_n x^{e_n} ,\,$ where $\, a_0 \!=\! 1, e_0 \!=\! 0. \,$ Then $\, f'(x) = \sum_{n=1}^\infty a_n e_n x^{e_n-1} ,\,$ and using the differential equation, $\, f(x) = \sum_{n=1}^\infty a_n e_n x^{2e_n-2}. \,$ Comparing the two series we get the recursion $\, e_{n+1} = 1 + e_n/2 \,$ with solution $\, e_n = 2 - 2^{1-n}. \,$ This solution for $\, e_n \,$ leads to the recursion $\, a_{n+1} = a_n / e_{n+1}. \,$ Since $\, e_n \to 2 \,$ as $\, n \to \infty, \,$ then $\, a_n \sim 2^{-n}. \,$ This implies that the Puiseux series for $\, f(x) \,$ converges for $\, 0 \le x \le 1. \,$ It also converges for $\, x>1 \,$ and its graph is given in the answer by Alejandro Salum. The general solution is multiplied by $\, f(0) \,$ and now if $\, f(0) = 0, \,$ then $\, f(x) = 0 \,$ for $\, 0 \le x. \,$ Besides this Puiseux series expansion, it connects with the answer by Christian Blatter which shows that it also has a power series expansion at $\, x=1, \,$ coming from $\, g(t):=f(e^t), \, $ where $\, g(t) \,$ has a power series expansion at $\, t=0. \,$ As a check $\, f(0) = 1, \, f(1) \approx 3.46274661945506, \, f(2) \approx 7.73614964618559.$

Note that this question asks about a kind of delay differential equation solution. A simple example is $\,f'(x) = f(x-1)\,$ where $\,f\,$ can be defined essentially arbitrarily on $\,0\le x\le 1\,$ and then for $\,x>1\,$ define $\, f(x) := f(1) + \int_0^{x-1} f(t)\, dt.\,$ Of course, if $\,f(x) = 0\,$ on $\,[0,1]\,$ then $\,f(x) = 0\,$ for all $\,x\ge 0.\,$ In the current question the D.D.E. is $\, f'(x) = f(\sqrt{x}).\,$ Following the simple example, we could define $\,f\,$ arbitrarily on $\,[c^{-2},c^{-1}]\,$ where $\,c>1\,$ is arbitary. The values of $\,f\,$ on $\,(0,c^{-2})\,$ are determined by $\, f(x) = f(c^{-2}) - \int_x^{c^{-2}} f(\sqrt{t})\,dt.\,$ The values for $\,f\,$ on $\,(c^{-1},1)\,$ is more demanding. We must have $\,f\,$ infinitely differentiable on $\,[c^{-2},c^{-1}]\,$ because $\,f(x) = f'(x^2) = 2x^2 f''(x^4) = 4x^2 f''(x^8) +8x^{10} f^{(3)}(x^8)\,$ and so on. In summary, if $\,f\,$ is infinitely differentiable on $\,[c^{-2},c^{-1}]\,$ then it can be uniquely extended to $\,(0,1).\,$

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  • $\begingroup$ Consider the function $g$ defined by $g(0) = 0$ and $g(x) = \exp(-1/x^2), x\ne0$. Then that function has derivatives of all orders at 0 and they are all zero, however it is not zero everywhere. $\endgroup$ – Quantaliinuxite Jun 2 '18 at 23:42
  • $\begingroup$ Wouldn’t you need to show the power series converges however? $\endgroup$ – Quantaliinuxite Jun 3 '18 at 6:58
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You can use chain rule to get:

$f'(x) = 2xf''(x^2)$

Now you have:

$f'(x^2) = 2 x^2 f''(x^4)$

Now substitute it on first equation:

$f(x) = 2 x^2 f''(x^4)$

And since $x \in [0,\infty)$ we have 3 cases:

(i) f(x) is constant $\Rightarrow$ $f(x) = f^{(n)}(x)=0$

(ii) f(x) is non decreasing $\Rightarrow$ $f^{(n)}(x)\geq0$, but in order to equations to hold $f^{(n)}(x)=0$; $n=0,1,...$

(iii)f(x) is non increasing $\Rightarrow$ $f^{(n)}(x)\leq0$, but in order to equations to hold $f^{(n)}(x)=0$; $n=0,1,...$

Concluding that $f(x) = 0$

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    $\begingroup$ The function doesn't have to have a second derivative $\endgroup$ – Jakobian May 31 '18 at 12:34
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    $\begingroup$ @Adam It does, since $f'(x)=f(x^{1/2})$ is the composition of two functions that are differentiable on $(0,\infty)$. $\endgroup$ – user565560 May 31 '18 at 12:35
  • $\begingroup$ Ok, on $(0, \infty)$. $\endgroup$ – Jakobian May 31 '18 at 12:36
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    $\begingroup$ @Arthur T Why are these three cases exhaustive? $\endgroup$ – Kavi Rama Murthy May 31 '18 at 12:43
  • $\begingroup$ Else we would have $f(x)$ and $f^{(n)}(x)$ with different signs for some $n$, which will be not possible because x in positive and if you keep using chain rule you will always have something like: $f(x) = c x^k f^{(n)}(x^l)$. $\endgroup$ – A.T May 31 '18 at 12:46

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