Finding all differentiable $f: [0,+\infty) \rightarrow [0,+\infty)$ such that $f(x) = f'(x^2)$ and $f(0)=0$

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 $f(0)=0$.

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 $[1, \infty)$: $$ e^{-x} f(x) \le e^{-1} f(1) = 0 \\ \implies f(x) \le 0 \implies f(x) = 0 \, . $$

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 $f$:

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.

$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$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, $f = 0$.