I'm reading a monograph that considers the following problem:
$$\inf_{z \in C^1} \int_0^1 c\left(\frac{dz(t)}{dt}\right) dt$$ $$z(0) = x, z(1) = y$$
Here $c$ is a convex function, $z(t)$ are paths with initial and final points given. They claim the infimum is $c(y-x)$ and this follows from Jensen's inequality. I can see part of it:
$$\int_0^1 c\bigg(\frac{dz(t)}{dt}\bigg) dt \geq c\bigg(\int_0^1 \frac{dz(t)}{dt} dt\bigg)=c(z(1)-z(0))=c(y-x)$$
They claim though that:
$$\inf_{z \in C^1} \int_0^1 c\left(\frac{dz(t)}{dt}\right) dt=c(y-x)$$
How do we know that the inf of LHS is the RHS -- can't the inf end up greater than the RHS?
Let $Z = \{z \in C^1([0,1], \mathbb{R}) : z(0) = x \text{ and } z(1) = y\}$.
You have shown that $c(y-x)$ is a lower bound of the operator $$\mathcal{F} : Z \to \mathbb{R} : z \mapsto \int_0^1 c \left(\frac{dz(t)}{dt} \right) dt$$
Now if we can show that $\mathcal{F}$ attains $c(y-x)$ at some function $z \in Z$, then $c(y-x)$ is the minimum of $\mathcal{F}$, consequently its infimum.
For that consider $z : [0,1] \to \mathbb{R} : t \mapsto (1-t)x + ty$. Thus $z \in Z$ and $$ \mathcal{F}(z) = \int_0^1 c \left(-x + y \right) dt = c \left(y - x \right) $$
This concludes your proof.