Fenchel conjugate of least squares in a Banach case

I can't wrap my head around this question: let $\mathcal{X}$ be a Banach space, let $\mathcal{X}^*$ be its dual and $\mathcal{H}$ be a Hilbert space (such as $L^2$). Define the linear mapping $\Phi : \mathcal{X} \to \mathcal{H}$ and the observation $y \in \mathcal{H}$. Then, for all $x \in \mathcal{X}$

$$ f(x) = || y - \Phi x ||_\mathcal{H}^2 $$

Is there a close form of the Fenchel conjugate of $f$, defined for all $x^* \in \mathcal{X}^*$ by:

$$ f^*(x^*) = \sup_{x \in \mathcal{X}} \langle x, x^*\rangle - f(x) $$

I tried Fermat rule but this does not reach any conclusive results.

Solution 1:

(Very) partial answer (too long for a comment): If ${\cal X, X^*, H}$ are all equal to $\mathbb R^n$ and, for simplicity $y=0$ and $\Phi$ is the identity. Then $f(x)=||x||^2$ is convex and the supremum is attained for $x$ satsifying $$ 0=x^*_i-\partial_if(x)=x^*_i-2x_i\quad\forall i=1,...,n\,. $$ Therefore, plugging $x=x^*/2$ into $\langle x,x^*\rangle-f(x)$ gives $$ f^*(x^*)=\sum_{i=1}^n\frac{(x^*_i)^2}{2}-f(x^*/2)=\frac{||x^*||}{2}-\frac{||x^*||}{4}=\frac{||x^*||}{4}\,. $$