I am considering an optimisation problem of the following form: $$\textrm{Minimise } \mathcal{C}_1[y] + \mathcal{C}_2[z] \textrm{ such that } z = y_x,$$ where $y$ and $z$ are functions of $x$, and $y_x$ denotes the derivative of $y$ with respect to $x$. $\mathcal{C}_1$ and $\mathcal{C_2}$ are functionals. We can form the Lagrangian for this minimisation problem: $$ \mathcal{L} = \mathcal{C}_1[y] + \mathcal{C}_2[z] - \langle \lambda, z - y_x \rangle$$ where $\lambda(x)$ is the Lagrange multiplier corresponding to the constraint $z=y_x$. Taking functional derivatives with respect to $y$ and $z$ gives: $$\frac{\delta \mathcal{L}}{\delta y} = \frac{\delta \mathcal{C}_1}{\delta y} - \lambda_x, $$ $$\frac{\delta \mathcal{L}}{\delta z} = \frac{\delta \mathcal{C}_2}{\delta z} - \lambda. $$ The optimality system (necessary condition for a minimiser) is that the above functional derivatives are zero along with the constraint $z=y_x$ being satisfied.
My question is how to place additional restrictions on the function space in which we are optimising. I know that for integral constraints, we can just add another Lagrange multiplier to account for the fact that any variations are also constrained.
Since $z = y_x$, we know that $y$ is more regular than $z$. Moreover, if $y$ is in a particular function space $Y$, we know that $z$ lives in the function space $Z$ of derivatives of the functions in $Y$ (e.g. if $Y$ is the space of polynomials of order $n$, then $Z$ is the space of polynomials of order $n-1$).
The sensitivity $\delta \mathcal{L}/\delta y$ is less regular than $\delta \mathcal{L}/\delta z$, with the former involving a $\lambda_x$ and the latter involving a $\lambda$. Moreover, if $\delta \mathcal{L}/\delta y$ is in $Y$, then it is not necessarily true that $\delta \mathcal{L}/\delta z$ is in $Z$ (e.g. if $\lambda_x$ is a polynomial of order $n$, then $\lambda$ will not necessarily live in the space of polynomials of order $n-1$).
For an easy example, let's set $\mathcal{C}_1[y] = \| y \|^2/2$ and $\mathcal{C}_2[z] = \| z \|^2/2$. We could throw away the Lagrange multiplier for $z = y_x$ and work directly with the objective $y^2/2 + y_x^2/2$, or proceed with the split form to end up with the equation to be solved by a minimizer, $y = y_{xx}$. But I want to take a step back and look at the functional derivatives: $$\frac{\delta \mathcal{L}}{\delta y} = y - \lambda_x $$ $$\frac{\delta \mathcal{L}}{\delta z} = z - \lambda $$ If I set $\lambda_x = y$ and solve for $\lambda$, then $z - \lambda$ might not live in $Z$.
I am not really sure how to deal with this kind of problem. I guess that since I have put $z=y_x$ into the Lagrangian, I have formally unconstrained the problem from this relationship. So maybe the best I can do is to project the variables $y$ and $z$ into the correct spaces.