Inf of Jensen's inequality

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 -- page 145, Prop 5.2)

• Just consider the straight line path with constant speed: $z(t)=(1-t)x+ty$. Feb 25 '18 at 3:48
• ya, plugging it in works. maybe i should have asked -- how can I see that's the solution. I tried using a direct method of calculus of variations and was getting mixed up. Feb 25 '18 at 5:07
• Jensen's inequality $\int c(f)\,d\mu \ge c(\int f\,d\mu)$ is always equality when $f$ is constant, so you would want $dz/dt$ to be constant to saturate. That means $z(t)$ has to be linear, and since you know the endpoints, you're done. Also useful to know is that when $c$ is strictly convex, a constant (almost everywhere) is the only way to saturate. Feb 25 '18 at 5:15
• aha! ya got it, add it as answer so I can accept! thx Feb 25 '18 at 5:18

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.