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)<F(u)-2\delta} and as a consequence for {n} great enough we have {F_{\varepsilon_n}(v_n)<F(u)-\delta}. 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}.

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  1. maxime
    September 5, 2017 at 4:53 pm

    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
    ( Iam french)

    • September 5, 2017 at 8:08 pm

      Sorry, but I have no idea how to help you on this one… You should write an exact description of your problem. I don’t even know what is a Yosida approximation…

  1. April 23, 2013 at 9:37 pm

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