I'm learning some Fenchel duality problems and I'm piecing together the logic of some problems that I have the work-through for. In both instances a similar approach is taken regarding the conjugate convex functional, but I can't seem to figure out or find out why it works. The first function is $$F(x)=\int_0^1 x(t)^2dt$$ with some given constraints, but the important part is that the convex conjugate is worked as: $$F^*(\zeta)=\sup\left(\int_0^1\zeta x-x^2dt\right)$$ Now, here the goal is to remove the $x$. This is done by $$F^*(\zeta)=\sup\left(\int_0^1\frac{1}{4}\zeta^2dt-\underbrace{\int_0^1(x-\frac{1}{2}\zeta)^2dt}_{=0}\right)$$ So, my question is how the underbraces term $=0$?
Again, in another the function is $$\int_0^1(x(t)-t)^2dt$$ With again some given constraints, but importantly the convex conjugate is manipulated as $$F^*(\zeta)=\sup\int_0^1\zeta x-(x-t)^2dt$$ And it seems the goal is the same as before, since we then have $$F^*(\zeta)=\sup\left(\int_0^1\underbrace{-(\frac{1}{2}\zeta-(x-t))^2}_{=0}+\frac{\zeta^2}{4}+\zeta tdt\right)$$ So, again same question, why is the underbraced term 0?
Thank you very much for any help.
The optimization variable is $x$. We have \begin{align} F^*(\zeta)=&\sup_x\left(\underbrace{\int_0^1\frac{1}{4}\zeta^2\,dt}_{\text{does not depend on }x}-\int_0^1(x-\frac{1}{2}\zeta)^2\,dt\right)=\int_0^1\frac{1}{4}\zeta^2\,dt+\sup_x\left(-\int_0^1(x-\frac{1}{2}\zeta)^2\,dt\right)=\\ =&\int_0^1\frac{1}{4}\zeta^2\,dt-\inf_x\left(\underbrace{\int_0^1(x-\frac{1}{2}\zeta)^2\,dt}_{\ge 0,\forall x\text{ and }=0,x=\zeta/2}\right)= \int_0^1\frac{1}{4}\zeta^2\,dt-0. \end{align} Similarly in the second case. It is always convenient to complete the square for optimization. By doing that we get a minimization of a non-negative function with the trivial minimum being zero.