Does there exist a minimizer in $C^1([0,1])$ (or $H^1([0,1])$) for $$\frac{1}{2}\int (f')^2 dx, \text{ given the boundary conditions: } f(0)=0, f(1) = a, f'(1) = b?$$
When $a=b$, we have the minimizer $f(x) = ax$.
But in general, how does the standard variational principle hold? Because when choosing test functions, we need $\phi(0) = \phi(1) = 0$ and $\phi'(1)=0$. But the variation is the same as normal Dirichlet condition, the $\phi'(1)=0$ part is lost $$\lim_{t\rightarrow 0} \frac{\frac{1}{2}\int (f'+t\phi')^2 dx - \frac{1}{2}\int (f')^2 dx}{t} = \int_0^1 f' \phi'dx$$ I get $f''\equiv 0$ which can not be true in general for the D and N the boundary conditions.
You have three conditions on the function but only one derivative in the integrand. In general you are already going to have an over conditioned problem.
Direct application of the Euler-Lagrange equation gives:$$\ddot{f} = 0$$
Which means, as you say, that the extreme function is linear. By inspection this is a minimizer for the functional. Your boundary conditions for $f$ completely determine it as you say. So $$f(x) = a x$$ is correct. This agrees with the condition on the derivative at $x = 1$ only when $a = b$.