Why does a local maximum point imply that the second derivative is zero in this problem? This is from chapter 6.1 in Introduction to Partial Differential equations (Tveito & Winther, 1st edition).
The book states roughly:
Let $v \in C^2((0,1)\cap C([0,1]))$ be a function satisfying the following inequality:
$v''(x) + a(x)v'(x) > 0, \ x \in (0,1)$,
where $a$ is a continuous function on $[0,1]$. Suppose $v$ has a local maximum in an interior point $x_0 \in (0,1)$. Then, we have
1) $v'(x_0) = 0 \ $ and
2) $v''(x_0) \leq 0$.
End quote.
The book then goes on to prove, by contradiction, that $v$ cannot have a local maximum on the interior.
My problem is that when reading https://mathworld.wolfram.com/SecondDerivativeTest.html, 1) seems like a necessary condition, while 2) seems like a sufficient one. Doesn't that mean that $v$ could have a local maximum at $x_0$, while at the same time $v''(x_0) > 0$?
 A: No, and the proof is straightforward. The second derivative test says if $f'(x_0)=0$ and $f''(x_0)<0$, then $f$ has a local maximum. If $f''(x_0)>0$, then $f$ has a local minimum.
If $f'(x_0)=0$ and $f''(x_0)>0$ and $f$ has a local maximum, then that means $(x_0,f(x_0))$ is both a local maximum and local minimum.
By definition of local maximum, there exists $\delta_1>0$ such that $f(x)\leq f(x_0)$ in $(x_0-\delta_1,x_0+\delta_1)$. By definition of local minimum, there exists $\delta_2>0$ such that $f(x)\geq f(x_0)$ in $(x_0-\delta_2,x_0+\delta_2)$. Hence, let $\delta=\min\{\delta_1,\delta_2\}$. Then $f(x)=f(x_0)$ in $(x_0-\delta,x_0+\delta)$.
Hence, $f$ is constant in a neighborhood so the second derivative is $0$, a contradiction.
I think your confusion is that the $<$ and $>$ parts are the sufficient parts. A function can fail the second derivative test and still have a local maximum. For example, let $f(x)=-x^4$ at $x_0=0$.
In the context of your two points, for a twice differentiable function both are necessary with strict inequality being sufficient.
