Second variation positive definite but not weak local minimum?

Question

Consider a functional $J \colon S \to \mathbb{R}$ where $S \subseteq C^2[a,b]$. Let $J(y) = \int_a^b f(x,y,y') \, dx$, let $y$ be an extremal (solution to the Euler-Lagrange equation) for $J$, and suppose the second variation of $J$ at $y$ is positive definite. Does it follow that $J$ has a weak local minimum at $y$, where $weak$ means with respect to the norm $$||y||_1 = \sup_{x \in [a,b]} |y(x)| + \sup_{x \in [a,b]} |y'(x)|?$$

Answer 1

I'll have to assume that $y$ is an inner point of $S$ and that there's some $\varepsilon >0$ so that in the ball $B_{\varepsilon} ( y ) \subset S$ one has $\delta^{2} J ( y;v ) >0$ for every $v \in B_{\varepsilon} ( y )$.

Fix some $v \in B_{\varepsilon} ( y )$ and define $\varphi ( t ) := y+t ( v-y )$ for $t \in [ 0,1 ]$. Then obviously $\varphi ( t ) \in B_{\varepsilon} ( y )$ for all $t$ and $\delta^{2} J ( \varphi ( t ) ;v-y ) >0$ too. Now define the new functional $I ( t ) := J ( \varphi ( t ) )$. This is twice differentiable if $J$ is and one has $I' ( t ) = \delta J ( \varphi ( t ) ,v-y )$ and $I'' ( t ) = \delta^{2} J ( \varphi ( t ) ;v-y ) >0$. Therefore, for some $\tau \in [ 0,1 ]$:

$$ I ( 1 ) -I ( 0 ) =I' ( 0 ) + \frac{1}{2} I'' ( \tau ) $$ and the first term is zero because $I' ( 0 ) = \delta I ( u,v-y )$ and the second is positive. This means

$$ J ( v ) -J ( y ) >0, $$

or: $y$ is a strict local minimizer of $J$.