Suppose $ F(a)=0, F(b)=0, a and $F''(x)<0$ for $a
Do we have $F(x) > 0$ for $a
Can you tell me an existing theorem that I can use in a paper to obtain the conclusion? I think there should be one, but I can't remember.
Suppose $ F(a)=0, F(b)=0, a and $F''(x)<0$ for $a
Do we have $F(x) > 0$ for $a
Can you tell me an existing theorem that I can use in a paper to obtain the conclusion? I think there should be one, but I can't remember.
This is a consequence of the convexity of $-F(x)$ on $[a,b]$.
Suppose $F(c) < 0$ for some $c \in (a,b)$. By the mean value theorem, there exist values $d_1 \in (a,c)$ and $d_2 \in (c,b)$ such that \begin{align} F'(d_1) = \frac{F(c) - F(a)}{c - a} = \frac{F(c)}{b-a} < 0 \\ ~ \\ F'(d_2) = \frac{F(b) - F(c)}{b - c} = \frac{-F(c)}{c-b}> 0, \end{align} and hence $F'(x)$ changes sign somewhere on $(a,b)$. This contradicts $F''(x) < 0$. Hence, $F(c) \geq 0$ for all $c \in (a,b)$.
Why not. can be concave/not and positive between any two points. the concavity of the curve may also be changing at x. Depends on the function.
There is no theorem that I know of that talks specifically about this case, but you can prove it using a bit of simple calculus.
How about this proof: By Rolle's theorem, $F'(x) = 0$ at least once in the interval $(a, b)$. However it must happen exactly once since $F''(x) < 0$... This critical point is a maximum point, so we have $F(x)$ ascending from $(a, 0)$ towards a maximum and then descending back to $(b, 0)$.