So, I have this differential equation problem to solve but only using real analysis The problem is:
Given f such that $f(0)=0$ and $f'(x) = [f(x)]^3$, show that $f(x) = 0\; \forall x \in [0,\infty)$. I wanted to use Cauchy-Picard, but my sister said it should be solved with basic real analysis only. Any help would be appreciated.
 A: Let $x_0$ be the suppremum of the $x$ such that $f(t)=0$ for $t\in[0,x]$.
If $x_0$ is finite: 
Let $1>a>0$ be such that $|f(x)|<1/M$ for your favorite $M>1$ on $[x_0,x_0+a]$.
Then $|f(x)|=\left|\int_{x_0}^{x}f'(t)dt\right|=\left|\int_{x_0}^{x}f^3(t)dt\right|\leq 1/M^3$ for $x\in[x_0,x_0+a]$.
Repeating the same argument, using the new found bound $1/M^3$ instead of $1/M$ and so on, we get that $|f(x)|<1/M^{3^{n}}$ for all $n$, on $x\in[x_0,x_0+a]$. Therefore $f(x)=0$ for $x\in[x_0,x_0+a]$.
Therefore $x_0$ is not finite.
A: Formally, this is a separable equation:
$$
{df \over dx} = f(x)^3,
$$
so:
$$
dx = {df \over f^3}.
$$
However, this would be valid only in a neighborhood of a state where $f(x) \neq 0$.  Since, by contrast, $f(0) = 0$, direct inspection shows that $f(x) \equiv 0$ is a solution.  To see that it is unique, consider the subsets of the $f$-axis invariant under the flow of the vector field $v(f) = f^{3}$.  There are only three such subsets: the strictly positive semiaxis $f>0$, the strictly negative semiaxis $f<0$, and the point $f=0$.  Since a solution to the ODE must be contained in one of the subsets, it follows that it must lie in $\{0\}$; i.e., $f \equiv 0$.
Source: [https://books.google.com/books?id=JUoyqlW7PZgC&printsec=frontcover&dq=ordinary+differential+equations&hl=en&sa=X&ved=0ahUKEwjWiomptPXUAhXEyT4KHWWMAvgQ6AEIKzAA#v=onepage&q=ordinary%20differential%20equations&f=false]
A: You could "unfold" the proof of Cauchy-Picard, in order to get a basic real analysis proof which would go something like this:
Since $f$ is differentiable, $f$ is continuous, so choose $\delta > 0$ such that $\delta \le 1$, and $|f(x)| < \frac{1}{2}$ whenever $|x| < \delta$.  Let $A := \sup \{ |f(x)| : x \in [0, \delta] \}$.  Then for $x \in [0, \delta]$, $f(x) = \int_0^x [f(t)]^3 \, dt$, so
$$|f(x)| \le \int_0^x |f(t)|^3 \, dt \le x A^3.$$
Therefore, $A \le \delta A^3 \le A^3$, which together with the fact that $0 \le A \le \frac{1}{2}$ implies that $A = 0$.
Therefore, we have shown that $f(x) \equiv 0$ for $x \in [0, \delta]$, for some $\delta > 0$.  From here, the proof concludes using the translation invariance of the differential equation, along with a connectedness type of argument.
A: Let $a=\min\left(1,\sup\left\{x\in\mathbb{R}^+: |f(x)|\leq 1\right\}\right)$.
Let us define $g(x)=\left\{\begin{array}{rcl} f(x) &\text{if} & x\in\left[0,\frac{a}{2}\right]\\ f(a-x)&\text{if}& x\in\left[\frac{a}{2},1\right]\end{array}\right.$ 
By applying Wirtinger's inequality to $g(x)$ we get:
$$\int_{0}^{a}g(x)^2\,dx \leq \frac{a^2}{\pi^2}\int_{0}^{a}g'(x)^2\,dx \leq \frac{a^2}{\pi^2}\int_{0}^{a} g(x)^6\,dx\leq \frac{a^2}{\pi^2}\int_{0}^{a}g(x)^2\,dx $$
hence $g(x)\equiv 0$ on $[0,a]$ and $f(x)\equiv 0$ on $\left[0,\frac{a}{2}\right]$. By shifting and repeating the same argument we get $f(x)\equiv 0$ on $\mathbb{R}^+$.
