I'm reading a monograph that considers the following problem:



> 
> $$\inf\_{z(t) \in C^1} \int\_0^1 c\bigg(\frac{dz(t)}{dt}\bigg) dt\\ z(0) = x, z(1) = y$$
> 
> 
> 


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:



> 
> $$\inf\_{z(t) \in C^1} \int\_0^1 c\bigg(\frac{dz(t)}{dt}\bigg) dt=c(y-x)$$
> 
> 
> 


How do we know that the inf of LHS is the RHS -- can't the inf end up greater than the RHS?


Edit:
As the commenter points out, a constant speed trajectory achieves the inf. I should ask: without knowing this answer ahead of time, I'm having trouble seeing how I could obtain this from this variational problem.


(Specifically, I am looking at this [book](http://people.math.gatech.edu/~gangbo/Cedric-Villani.pdf) -- page 145, Prop 5.2)



Inf of Jensen's inequality
Jensen's inequality $\int c(f)\,d\mu \ge c(\int f\,d\mu)$ is always equality when $f$ is constant, so you know that taking $dz/dt$ constant would saturate it. That means $z(t)$ has to be linear, and since you know the endpoints, $z$ is determined. 


Also useful to know is that when $c$ is strictly convex, a constant (almost everywhere) is the only way to saturate, so in that case no other function would do.


