7
$\begingroup$

I'm reading a proof on the existence of a solution to a minimisation problem, but I'm stuck. I give a brief summary of the arguments up to the point at which I'm stuck(at the yellow box).

Information:

The feasible set $X$ is sequentially compact with respect to weak$^\star$ convergence, and the objective function $f: X \longrightarrow \mathbb{R}$ is nonnegative.

The arguments:

We consider the set

\begin{align} W = \{ f(x): x \in X \}. \end{align}

Now, since $f$ is nonnegative, $W$ is bounded from below and therefore has an infimum. Write $\kappa = \inf W$. Then, by the definition of the infimum there is a sequence $(f(x_n))_n \subset W$ such that

\begin{align} f(x_n) \rightarrow \kappa. \end{align}

By sequential compactness we can find a convergent subsequence $(x_{n_k})_k$ of $(x_n)_n$, such that

\begin{align} x_{n_k}\overset{w^\star}{\rightarrow}x^\star \end{align}

for some $x \in X$. Clearly, we also have that

\begin{align} f(x_{n_k}) \rightarrow \kappa. \end{align}

Hence, the proof is completed by showing that the functional $f$ is lower semicontinuous, that is, if $(x_n)_n$ is convergent with weak$^\star$ limit $x$, then \begin{align} \lim\inf\limits_{n \rightarrow \infty} f(x_n) \geq f(x). \end{align}

My problem:

I don't see why the argument if the yellow box works. I guess I am confused about what is meant by existence here. I thought that existence of a solution was defined as follows: There is an $x^\star \in X$ such that

\begin{align} \inf_{x \in X}f(x) = f(x^\star). \end{align}

I do not see how the lower semicontinuity of $f$ implies this.

More information:

  • $X$ is convex.
$\endgroup$
  • $\begingroup$ Briefly, after you have got that $f(x_n)\to\kappa$ and $x_n\to x^\star$ what's left is to ensure that $f(x^\star)=\kappa$, i.e. no jump occurs when taking the limit. $\endgroup$ – A.Γ. Jul 12 '15 at 22:45
3
$\begingroup$

The point of lower semicontinuity (or something like it) is that were it to fail, there's no guarantee that $f(x^\ast)$ has the right value.

For example think about what goes wrong in the following: consider $X=[-2,2]$ and the function $f$ which takes $x$ to $-x$ for $x<0$ and takes $x$ to $x+1$ to $x\geq 0$. Take a minimizing sequence $x_n\in[-2,2]$ which of course will converge to $x^\ast=0.$ But $f(0)=1$.

The problem is that on a completely a priori level, having only some information about $X$ and some weak information about $f$ such as nonnegativity, there's no connection between doing something entirely within $X$ (such as taking a weak limit) and doing something with the corresponding values of $f$. (This is the problem above.)

But the assumption of lower semicontinuity is enough to establish a link. Taking only sequentially compact $X$ and nonnegative $f$, as you said we have a minimizing sequence $x_n$ with a weak limit $x^\ast$. And then there's nothing to say about $f(x^\ast)$ except that it's nonnegative. But now assuming semicontinuity, there is $f(x^\ast)\leq\liminf f(x_n)$. But $x_n$ was defined in such a way that $\liminf f(x_n)$ is the minimal possible value that $f$ takes on $X$. And the lower semicontinuity says $f(x^\ast)$ is less than or equal to that value! So $f(x^\ast)$ has to be exactly that value.

$\endgroup$
  • $\begingroup$ The last section was enough - thank you very much! $\endgroup$ – northwiz Jul 12 '15 at 22:45
  • $\begingroup$ @youler this is a very nice answer, where did you go youler? $\endgroup$ – yoshi May 28 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.