Perturbation Theory and $\delta J(v) = \langle DJ(v) \; , \; \delta v \rangle$? In my optimization course, we are talking about using perturbation theory on bilinear functionals and stuff. We were told to use the following relationship to describe how the perturbation $\delta$ in the variable $v$ interacts with the perturbation in the functional $J$: $$\delta J(v) = \langle DJ(v) \; , \; \delta v \rangle$$ So the perturbation of $J$ is equal to the inner product of the perturbation of $v$ and the differential of $J$ at $v$. I can use this just fine, however I don't feel like I get where it comes from; we were just given this statement as a fact (or I missed the explanation). How does this equivalence come about?

One way to define the gradient $DJ$ of a vector function $J(v)$ is according to the approximation

$$J(v+\delta v)=J(v)+\langle DJ(v),\delta v\rangle + o(\delta v),$$ where $o(x)$ denotes a function that tends to $0$ as $x\to 0$.

The notation $\delta J(v)$ suggests a small change in the function $J(v)$ of size $\delta v$, that is, $$ \delta J(v)=J(v+\delta v)-J(v). $$ Combining these two equations and discarding the $o(\delta v)$ term (which is asymptotically negligible in magnitude compared to $\delta v$) yields the equation you wrote.