Given the functional $F:H^1(0,1)\cap \left\{f(0)=a, f(1)=b\right\} \rightarrow \mathbb{C}$
such that $$F(g):= \int_0^1 g(x)f(x)dx.$$
where $f$ is a fixed $H^1$-function. I would like to find an upper bound for the minimizer of this function in $H^1$-norm such that $F(g)=c$ where $a,b,c$ are fixed real constants.
If anything is unclear, please let me know.
You can actually find the minimizer using Lagrange multipliers (in the infinite dimensional case) Lagrange for Banach. Miniminizing $\Vert g\Vert_{H^1}$ is the same as minimizing $\Vert g\Vert_{H^1}^2$, so using Lagrange multipliers, you are finding critical points of $$ H(g)=\Vert g\Vert_{H^1}^2+\lambda \left(\int_0^1 fg\,dx-c\right).$$ If you differentiate, that is, given $h\in H^1_0$ you consider the derivative of the function of one variable $p(t)=H(g+th)$, you find $$0=p'(0)=\int_0^1(2g'h'+2gh+\lambda fh)\,dx.$$ If $g$ is smooth, then you can integrate by parts to get $$0=\int_0^1(-2g''+2g+\lambda f)h\,dx.$$ Since this is true for all $h$ you find that $$-2g''+2g+\lambda f=0.$$ This is a linear differential equation. You can solve this using variations of parameters (see here second order ode)