Constant Speed of Geodesics Let V be the set of smooth functions $ f : [0,1] \rightarrow \Bbb R $ such that $ \int_0^1 f(t) dt = k $. If $ F : V \rightarrow \Bbb R $ is given by $ F(f) = \int_0^1 f(t)^2 dt $, then show that the only critical point of F is the constant function $ f(t) = k$. Deduce that geodesics have constant speed.
This looks like it should be pretty straight forward, but I can't manage it. I someone able to give me a (fairly sizeable!) hint, but not just a full solution. (If I still can't get it after a hint, then I may ask for a full solution!)
(I can show that $ f(t) = k $ is a critical point, but not that it is the only critical point.)
Thanks in advance! :)

I'd set this up as a constrained optimization problem with Lagrange multipliers: 
$$\mathcal{L}(f,\lambda) = \int_0^1 (f(t))^2\ dt + \lambda\bigg( \int_0^1 f(t)\ dt - k\bigg).$$
Now if you take a compactly supported perturbation $\epsilon\phi$ and differentiate at $\epsilon = 0$, a critical point must satisfy 
$$0 = \frac{d}{d\epsilon}\bigg|_{\epsilon = 0} \mathcal{L}(f+\epsilon\phi,\lambda) = \cdots$$
What Euler-Lagrange equation must $f(t)$ satisfy? 
To finish the problem, since you know $\int_0^1f(t)\ dt = k$, what can you conclude about $f$?
