$\Gamma$-convergence is a notion of convergence for functionals which has the nice property that if $x_\varepsilon$ are minimisers for a family of functionals $\{F_\varepsilon, \varepsilon > 0\}$ which $\Gamma$-converge to $F$, then $x_\varepsilon \rightarrow x$ implies that $x$ is a minimiser for $F$.

In the context of optimal control and calculus of variations type problems, the most common example I've seen is to use $\Gamma$-convergence to handle a $1/\varepsilon$ scaling in the Lagrangian. Are there any good examples of $\Gamma$-convergence being used when $\varepsilon$ is a parameter which determines the domain, perhaps through a state constraint?

Here is a toy problem to illustrate what I mean. Let $\mathcal{A}_{\varepsilon} (y, T)$ denote the class of absolutely continuous functions $x : [0, T] \rightarrow [\varepsilon, \infty)$ which satisfy $$ x (0) = \varepsilon, \quad x (T) = y. $$ For $x \in \mathcal{A}_{\varepsilon} (y, T)$, define the functional $F_{\varepsilon} (x)$ by $$ F_{\varepsilon} (x) = \frac{1}{2} \int_0^T \| \dot{x} (s) \|^2 ds. $$ Clearly, the minimiser $x_{\varepsilon} \in \mathcal{A}_{\varepsilon} (y, T)$ is $x_{\varepsilon}(t) = \frac{y - \varepsilon}{T} t + \varepsilon$, and $F_{\varepsilon} (x) = \frac{1}{2} \frac{(y - \varepsilon)^2}{T}$.

Does the sequence $\{ F_{\varepsilon}, \varepsilon > 0 \}$ $\Gamma$-converge to $F$, where $F$ is the functional which requires $x (0) = 0$? I would imagine so, modulo setting up the problem correctly. For instance, all the $F_\varepsilon$ should be defined on the same topological space, i.e., the same space of functions. There may be multiple way to address this, for instance, extending the definition of $F_\varepsilon$ to $C([0,T])$ by setting $F_\varepsilon(x) = +\infty$ for non-admissible $x$. Has anyone sorted out what the right approach is?

  • 1
    $\begingroup$ Did you ever find an answer to this question? It's extremely interesting to me. $\endgroup$
    – ABIM
    Commented Aug 23, 2018 at 16:05
  • $\begingroup$ @AIM_BLB I did not, though I convinced myself that it would for the example described above. $\endgroup$
    – snar
    Commented Aug 23, 2018 at 19:02
  • $\begingroup$ Oh ya? What was the reasoning? $\endgroup$
    – ABIM
    Commented Aug 24, 2018 at 22:20
  • $\begingroup$ To talk about $\Gamma$-convergence you need to define your functionals on the same space. If the natural spaces for $F_\varepsilon$ change with $\varepsilon$ you use $+\infty$ for elements outside your domain. $\endgroup$ Commented Sep 2, 2022 at 20:12

1 Answer 1


You wrote:

"There may be multiple way to address this, for instance, extending the definition of $F_\varepsilon$ to $C([0,T])$ by setting $F_\varepsilon(x) = +\infty$ for non-admissible $x$."

You can do slightly better: set $F_{n}=+\infty$ for $x(0)> \frac 1n$. Clearly it is "convenient" to pick the largest available value for $x(0)$ (as soon as $\frac 1n<y$).

This way your sequence $F_n$ is (from some $\bar{n}$ on) pointwise non-decreasing and you can use "easy" Gamma-convergence arguments like the ones in Section 1.8.1 in this nice reference book.(Keep in mind that sometimes the author uses increasing and decreasing for non-decreasing and non-increasing).


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