Derivatives of first/second order going to zero

Question

I have a smooth function such that $F(0)=F(h_1)=F(h_2)=0$ for some $h_1,h_2$ fixed ($F$ depends of $h_1,h_2$).

I also have that $F'(0)=a,F''(0)=b$.

I need to prove that if $h_1,h_2\rightarrow 0$, so $a,b\rightarrow 0$.

Well, $a=0$ by the Mean Value Theorem, because there's $c\in [0,h_1]$ s.t. $F'(c)=0$. As $h_1\rightarrow 0, c\rightarrow 0$ and $F'(c)\rightarrow a$. So, $a\rightarrow 0$.

Is it OK?

But what about the second derivative?

Many thanks!

tag mean-curvature-flows doesn't necessarily have much to do with Mean Value Theorem — Mirko, Sep 12 '19 at 01:30
Thanks, I retagged to limits, I do not know if this is better... — Quiet_waters, Sep 12 '19 at 01:31
How do you know $F'(c) \to a$? Do you assume continuity of $F'$? — kccu, Sep 12 '19 at 01:37
you are welcome, I didn't do much, glad someone answered your question and you accepted the answer — Mirko, Sep 12 '19 at 01:54

score 1 · Accepted Answer · answered Sep 12 '19 at 01:36

1

Sketch:

Suppose wlog that $h_1<h_2$. Then, by the mean value theorem, there is some $d$ in $(h_1,h_2)$ such that $F'(d)=0$. Then, by the mean value theorem, there exists some $e$ in $(c,d)$ such that $(F')'(e)=0$. Now, use your original argument.

answered Sep 12 '19 at 01:36

Michael Burr

33,866

Many thanks for the help, I could get it. – Quiet_waters Sep 12 '19 at 01:51

Derivatives of first/second order going to zero

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