## A brief introduction to Gamma-convergence

1. A motivating example

Consider the $p$-Laplace equation

$\text{div}(|Du|^{p-2}Du) = 0$ in $U$,

where $U$ is a nonempty bounded open subset of $\mathbb{R}$, and $u=g$ on $\partial U$. The energy functional associated with the PDE is

$I[w] = \int_U |Dw|^p dx$.

For $p>1$, because of convexity, $I[\cdot]$ has a unique minimizer $u^{(p)}\in W^{1,p}(U)$ which solves the $p$-Laplace equation.

Now, suppose that $u^{(p)}\rightarrow u^{(\infty)}$ for some function $u^{(\infty)}$ as $p\rightarrow\infty$ (in certain sense). Does $u^{(\infty)}$ solve any PDE?

First note that by some computations, for $p>1$,

$\displaystyle\text{div}(|Du|^{p-2}Du) = |Du|^{p-4}\left[|Du|^2\Delta{u} + (p-2) \sum_{i,j=1}^n u_{x_ix_j}u_{x_i}u_{x_j}\right] = 0$.

Formally, $u^{(p)}$ solves $|Du|^2\Delta{u} + (p-2) \sum_{i,j=1}^n u_{x_ix_j}u_{x_i}u_{x_j} = 0$. Dividing both sides by $p-2$, the equation becomes

$\displaystyle \frac{1}{p-2}|Du|^2\Delta{u} + \sum_{i,j=1}^n u_{x_ix_j}u_{x_i}u_{x_j} = 0$.

Letting $p\rightarrow\infty$, $u^{(\infty)}$ should solve $\Delta_{\infty}u := \sum_{i,j=1}^n u_{x_ix_j}u_{x_i}u_{x_j} = 0$.

Moreover, if we define $I_p[w] = \frac{1}{|U|} \int_U |Dw|^p$ (which is essentially the same as $I$ above), then $(I_p[u])^{1/p}\rightarrow \|Du\|_{L^\infty(U)}=:I_\infty[u]$ as $p\rightarrow\infty$. So formally $u^{(p)}$ converges to a minimizer of $I_\infty[\cdot]$. But does the above discussion really make sense? For example, $(I_p)^{1/p}\rightarrow I_\infty$ might make no sense because the domains of $I_p$‘s are different!

To rigor above discussion, our goal is to develop a theory of functional convergence such that

1. $u^{(p)}$ converges to $u^{(\infty)}$ which is a minimizer of $I_\infty$; and even better (if possible)
2. $u^{(\infty)}$ solves a “limiting” equation.

We will do 1 only. For 2, one may refer to the paper “Extension of functions satisfying Lipschitz conditions” by Gunnar Aronssov.

2. $\Gamma$-convergence

Definition ($\Gamma$-convergence). Let $M_\varepsilon$ and $\mathcal{N}$ be a family of topological spaces. A family of functionals $E_\varepsilon$ (defined on $M_\varepsilon$) $\Gamma$-converges to a functional $F$ (defined on $\mathcal{N}$) if the following hold:

1. (Compactness) If $E_\varepsilon(u_\varepsilon)\leq C$, where $u_\varepsilon\in M_\varepsilon$, then up to extraction, $u_\varepsilon\rightarrow u$ for some $u\in\mathcal{N}$.
2. (Lower bound) If $v_\varepsilon\rightarrow v$ for some $v\in\mathcal{N}$ ($v_\varepsilon\in M_\varepsilon$), then $\displaystyle \liminf_{\varepsilon\rightarrow 0} E_\varepsilon(v_\varepsilon) \geq F(v)$.
3. (Recovery sequence) For any $w\in\mathcal{N}$, there exist $w_\varepsilon\in M_\varepsilon$ such that $w_\varepsilon\rightarrow w$ and $\displaystyle \limsup_{\varepsilon\rightarrow 0} E_\varepsilon(w_\varepsilon) \leq F(w)$.

Note that in the above definition, we do not specify what is meant by $\rightarrow$. It has to be specified in each application. Also, we may replace the $\leq$ in 3 by $=$ because of 2.

Now we prove the fundamental theorem, and probably the only theorem of $\Gamma$-convergence:

Theorem 1. If $E_\varepsilon$ $\Gamma$-converges to $F$ and if $u_\varepsilon$ is a minimizer of $E_\varepsilon$ with $E_\varepsilon(u_\varepsilon)\leq C$, then up to extraction, $u_\varepsilon\rightarrow u\in\mathcal{N}$ and $u$ is a minimizer of $F$.

Proof. By compactness, $u_\varepsilon\rightarrow u\in\mathcal{N}$ up to extraction. By lower bound,

$\displaystyle \liminf_{\varepsilon\rightarrow 0}E_\varepsilon(u_\varepsilon) \geq F(u)$.

We claim that $u$ is a minimizer of $F$. Let $w\in\mathcal{N}$. Then there exists a recovery sequence $w_\varepsilon\in M_\varepsilon$ such that $w_\varepsilon\rightarrow w$ and $\displaystyle\limsup_{\varepsilon\rightarrow 0} E_\varepsilon(w_\varepsilon)\leq F(w)$. Since $u_\varepsilon$ minimizes $E_\varepsilon$, $E_\varepsilon(u_\varepsilon)\leq E_\varepsilon(w_\varepsilon)$ and so

$\displaystyle F(u)\leq \liminf_{\varepsilon\rightarrow 0} E_\varepsilon(u_\varepsilon) \leq \limsup_{\varepsilon\rightarrow 0} E_\varepsilon(w_\varepsilon) \leq F(w)$.

Therefore $u$ minimizes $F$. $\Box$

3. Returning to our example

Now we go back to our example in the first section and see how $\Gamma$-convergence applies.

Let $g$ be a Lipschitz function on $\partial U$. We define the following:

• $E_p = I_p^{1/p}$ defined on $M_p = \{u\in W^{1,p}(U): u=g\text{ on }\partial U\}$.
• $F=I_\infty$ defined on $\mathcal{N}=\{u\in W^{1,\infty}(U): u=g\text{ on }\partial U\}$.
• $\rightarrow$ means the weak convergence in $W^{1,q}(U)$ for all $1.

With the above definitions, we obtain

Theorem 2. $E_p$ $\Gamma$-converges to $F$ as $p\rightarrow\infty$.

Therefore, by Theorem 1, $u^{(p)}$ converges to a minimizer $u^{(\infty)}$ of $I_\infty$.

Proof. The recovery sequence is very easy to construct. If $w\in\mathcal{N}$, then $w\in M_p$ for all $p>1$ (recall that the domain $U$ is bounded). Therefore, we can take the recovery sequence to be just the constant sequence $w$.

We now prove the compactness. Suppose $u_p\in M_p$ with $E_p(u_p)\leq C$. Fix $1. By Hölder’s inequality, if $p\geq q$,

$\displaystyle\|Du_p\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \|Du_p\|_{L^p(U)}\frac{1}{|U|^{1/p}} = E_p(u_p)\leq C$.

Therefore, $\{u_p\}_{p\geq q}$ is bounded in $W^{1,q}(U)$ (you might need to think carefully why). Thus, we can extract a subsequence, still called $\{u_p\}$ for simplicity, converging weakly to $u\in W^{1,q}(U)$ with $u=g$ on $\partial U$. We can repeat this argument for $q+1$, $q+2$, $\ldots$ so that in each step, we select the weakly convergent sequence as a subsequence of the previous sequence. Taking the diagonal sequence, we conclude that there exists a subsequence $\{u_{p_j}\}$ such that $u_{p_j}\rightharpoonup u$ in $W^{1,q}(U)$ for all $1. This implies $u\in W^{1,q}(U)$ for all $1. Now, by lower semicontinuity,

$\displaystyle\|Du\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \liminf_{p\rightarrow\infty}\|Du_p\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \liminf_{p\rightarrow\infty} E_p(u_p)\leq C$.

Taking $q\rightarrow\infty$, $\|Du\|_{L^\infty(U)}\leq C$. That is, $u\in\mathcal{N}$.

It remains to show the lower bound condition holds. Suppose $v_p\in M_p$ such that $v_p\rightarrow v$ for some $v\in\mathcal{N}$. That is, $v_p\rightharpoonup v$ in $W^{1,q}(U)$ for all $1. Then same argument as before shows

$\displaystyle\|Dv\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \liminf_{p\rightarrow\infty}\|Dv_p\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \liminf_{p\rightarrow\infty} E_p(v_p)$.

Let $q\rightarrow\infty$, we get $\displaystyle F(v)\leq \liminf_{p\rightarrow\infty} E_p(v_p)$. $\Box$