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 $[0,1]$.


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 :-)

  • $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – Martin R Feb 6 '19 at 12:10
  • $\begingroup$ I have edited my question and solved the first condition,but I can't solve the second question.I hope I can get some help. $\endgroup$ – LiTaichi Feb 15 '19 at 12:24
  • $\begingroup$ Where does the question come from? $\endgroup$ – punctured dusk Feb 15 '19 at 12:31
  • $\begingroup$ a contest question of my univerisity.I have no idea about it. $\endgroup$ – LiTaichi Feb 15 '19 at 13:18
  • $\begingroup$ @ Martin R Could you please unclose this question? I have edited this. $\endgroup$ – LiTaichi Feb 16 '19 at 2:35

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.

  • $\begingroup$ So my question is $C=1$,your answer is exactly what I need! $\endgroup$ – LiTaichi Feb 20 '19 at 9:22

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