Home > Calculus of Variations > The Basic properties of Gamma Convergence

## The Basic properties of Gamma Convergence

Let ${X}$ be a metric space, and for ${\varepsilon >0}$ let be given ${F_\varepsilon : X \rightarrow [0,\infty]}$. We say that ${F_\varepsilon}$ ${\Gamma}$-converges to ${F}$ on ${X}$ as ${\varepsilon \rightarrow 0}$, and we write ${\Gamma-\lim F_\varepsilon =F}$, if the following conditions hold:

(LI) For every ${u \in X}$ and every sequence ${(u_\varepsilon)}$ such that ${u_\varepsilon \rightarrow u}$ in ${X}$ we have

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

(LS) For every ${u \in X}$ there exists a sequence ${(u_\varepsilon)}$ such that ${u_\varepsilon \rightarrow u}$ in ${X}$ and

$\displaystyle \limsup_{\varepsilon \rightarrow 0}F_\varepsilon(u_\varepsilon)\leq F(u).$

The ${\Gamma}$-convergence has the following properties:

1. The ${\Gamma}$-limit ${F}$ is always lower semicontinuous on ${X}$;

2. ${\Gamma}$-convergence is stable under continuous perturbations: if ${F_\varepsilon \stackrel{\Gamma}{\longrightarrow} F}$ and ${G}$ is continuous, then

$\displaystyle F_\varepsilon + G \stackrel{\Gamma}{\longrightarrow} F+G$

3. If ${F_\varepsilon \stackrel{\Gamma}{\longrightarrow} F}$ and ${v_\varepsilon}$ minimizes ${F_\varepsilon}$ over ${X}$, then every limit point of ${(v_\varepsilon)}$ minimizes ${F}$ over ${X}$.

I have seen these properties stated in many places, but the proofs are usually left to the reader. I will try and give the proofs below.

Proofs 1. Take ${u_n}$ a sequence in ${X}$ such that ${u_n \rightarrow u}$ in ${X}$ and ${\varepsilon_n \rightarrow 0}$. Assume that there exists ${\delta>0}$ for which we have ${\liminf\limits_{n \rightarrow \infty} F_{\varepsilon_n}(u_n) \leq F(u)-3\delta}$.

The properties ${(LI),(LS)}$ of the ${\Gamma}$ convergence combined prove that for every ${v \in X}$ there exists a sequence ${v_\varepsilon \rightarrow v}$ in ${X}$ such that ${\lim\limits_{n \rightarrow \infty} F_\varepsilon(v_\varepsilon) = F(v)}$.

We apply this result to each of the ${u_n}$ above and we find that there exists a sequence ${v_n}$ such that ${d(u_n,v_n)<1/n}$ and ${|F_{\varepsilon_n}(v_n)-F(u_n)|<\delta}$. Obviously we have ${v_n \rightarrow u}$ and by the ${(LI)}$ property of the ${\Gamma}$ convergence

$\displaystyle \liminf_{n \rightarrow \infty} F_{\varepsilon_n} (v_n) \geq F(u).$

On the other hand for ${n}$ great enough we have ${F_{\varepsilon_n}(u_n) and as a consequence for ${n}$ great enough we have ${F_{\varepsilon_n}(v_n). We have reached a contradiction. Therefore we have

$\displaystyle \liminf_{n \rightarrow \infty} F(u_n) \geq F(u)$

and ${F}$ is lower semicontinuous.

2. It is easy to see that ${G}$ verifies both ${(LI),(LS)}$ with equality, for every sequence ${u_\varepsilon \rightarrow u}$, thus proving the desired result immediatley by using the properties of ${\liminf, \limsup}$ for the sum of two sequences.

3. Suppose there exists ${v}$ and ${(\varepsilon_n)\rightarrow 0}$ such that ${v_{\varepsilon_n} \rightarrow v}$. Take another point ${w \in X}$ and the recovery sequence (which exists as a consequence of the ${\Gamma}$-convergence) ${w_n \rightarrow w}$ such that

$\displaystyle \lim_{n \rightarrow \infty} F_{\varepsilon_n}(w_n)=F(w).$

Then we have

$\displaystyle F(w)=\lim_{n \rightarrow \infty} F_{\varepsilon_n}(w_n) \geq \liminf_{n \rightarrow \infty} F_{\varepsilon_n}(v_{\varepsilon_n})\stackrel{\text{(LI)}}{\geq} F(v)$

proving that ${F(w) \geq F(v)}$. Since this is true for every ${w \in X}$ we find that ${v}$ is a minimizer for ${F}$.

please can you help me to prouve that $q(x)\leq \liminf_{n\rightarrow +\infty}q_n(x_n)$
where $q(x)=\,$, $q_n(x_n)=\,$, A an operator in Hilbert space H ans $A_n$ the approximation Yosida of A and (x_n) sequence witch converge to x in H