$f^{(n)}\leq1+|f|+|f'|+...+|f^{(n-1)}|$,Prove that $f$ has upper bound. If$f\in C^n[0,1)$ such that$$f^{(n)}\leq1+|f|+|f'|+...+|f^{(n-1)}|$$
for all n$\in\mathbb{N^+}$
Prove:$f$ has a upper bound
My attempt:
We can get $f^{(n)}\leq2^{(n-1)}(1+f)$ from the inequality.so by Taylor's theorem.
$$f(x)\leq\frac{1+f(0)}{2}(\frac{2x}{1!}+\frac{2^2x^2}{2!}+...+\frac{2^{n}x^n}{n!})+O(x^n)$$
so,the $f$ has upper bound in $[0,1]$.
$MY\quad QUESTION$
If the inequality isn't true for all $n\in \mathbb{N^+}$,that is the inequality is only true for $n(k<n,k\in\mathbb{N^+}$it isn't true),Does the $f$ has a upper bound in $[0,1]$?
But the details of the proof are killing me,any help would be greatly appreciated :-)
 A: Here is a suggestion for a strategy for this problem. It's not a complete answer, and I suspect it's much more complicated than necessary but I don't see a slicker argument.
For the purposes of induction, it will help to aim for a slightly stronger statement:

For $n\geq 0$ and $f\in C^n[0,1),$ if there exists $C>0$ such that $f^{(n)}\leq C(1+|f|+\dots+|f^{(n-1)}|)$ then $f,f^{(1)},\dots,f^{(n)}$ are bounded above.

The base case $n=0$ is obvious. For $n>1$ this can be split into two cases.
Case 1: $\liminf_{x\to 1} f^{(n-1)}/(1+|f|+\dots+|f^{(n-2)}|)>-\infty$
In this case, we can apply the induction hypothesis to $-f^{(n-1)}$ to show that $f,\dots,f^{(n-1)}$ are bounded below, say $f^{(i)}\geq -B$ for some $B>0.$ This implies $|f^{(i)}|\leq f^{(i)}+B$ for $i=0,\dots,n-1.$ So $f^{(n)}\leq C'(1+f+\dots+f^{(n-1)})$ with $C'=1+nB.$ It should be possible to bound $f$ by comparing to a nonhomogeneous linear ODE, or to a function of the form $Ae^{Dx}.$ At least we've got rid of the absolute value signs. And $g=f'$ satisfies $g^{(n-1)}\leq C'(1+|g|+\dots+|g^{(n-2)}|)$ for some $C',$ so we can bound $g,$ and so on, so all of $f,\dots,f^{(n)}$ should be bounded.
Case 2:$\liminf_{x\to 1} f^{(n-1)}/(1+|f|+\dots+|f^{(n-2)}|)=-\infty$
Write $H(x)=-f^{(n-1)}$ and $G(x)=1+|f|+\dots+|f^{(n-2)}|.$
Assuming $H(x)>0,$ we have $H'(x)=-f^{(n)}(x)\geq -C(H(x)+G(x)).$ Also, I want to say $G'(x)\leq H(x)+G(x).$ This is not quite rigorous since $G$ may not be differentiable, but it should be possible to perform the following comparison argument anyway.
As long as $H(x)>0$ holds we get
$$(1+\tfrac HG)'=(\tfrac HG)'=(H'G-G'H)/G^2\geq (-CHG-CGG-HH-GH)/G^2\geq -C(1+\tfrac HG)^2$$
assuming wlog $C>1.$ 
We can therefore compare $1+\tfrac HG$ to solutions to the Riccati equation $y'+Cy^2=0.$ These are of the form $y=\tfrac{1}{(x/C)-T}.$ In particular we know there exists $x\in(0,1)$ with $1+\tfrac{H(x)}{G(x)}>C/x$ which gives $1+\tfrac{H(x')}{G(x')}>C/x'>1$ for all $x'\in(x,1).$
So $H(x)/G(x)$ is bounded below, and by induction $f,\dots,f^{(n-1)}$ are bounded above, which also means $f^{(n)}$ is bounded above.
