If $f$ is twice differentiable real function on $(0,∞)$ and $a,b,c $ are supremum of $|f|,|f'|$ and $|f''|$ on $(0,∞)$, then --

Asked Feb 05 '20 at 13:04

Active Feb 05 '20 at 13:04

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If $f$ is twice differentiable real function on $(0,∞)$ and $a,b,c $ are supremum of $|f|,|f'|$ and $|f''|$ on $(0,∞)$ then

$a)$. $ b^{2} ≤ ac$

$b)$ $ b^{2} < ac$

$c)$. $ b^{2} = ac$

$d)$. $ b^{2} > ac$

I took $f(x) = \tan^{-1} (x)$, then discarded $c,d$. I am confused in between $a$ and $b$. Any hint$?$

asked Feb 05 '20 at 13:04

currently_into_ai

Possibly related: Prove $\sup \left| f'\left( x\right) \right| ^{2}\leqslant 4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right|$. – According to that Q&A you have $b^2 \le 4ac$. Could it be that a factor $4$ is missing in your question? – Martin R Feb 05 '20 at 13:48
1

In the linked question it is commented that the stated inequality is wrong (e.g., for $f(x):=x$) without further assumptions. – Christian Blatter Feb 05 '20 at 13:55

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