Lower hemicontinuity of intersection of relations

Question

This question is about the proof of a Lemma 7.1 of the paper

E. Michael, Continuous selections I.,Annals of Mathematics Second Series, Vol. 63, No. 2 (Mar., 1956), pp. 361-382 (22 pages) Published By: Mathematics Department, Princeton University

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Statement of the problem:

Suppose $\phi$ is a lower hemicontinuous relation from $X$ to $Y$ such that $\overline{\phi(x)}=\phi(x)\neq\emptyset$ for each $x\in X$. For each $x\in X$ let $$ m(x)=\inf\{\|y\|:y\in\phi(x)\}$$ Assume there is a lower semicontinuous function $\rho:X\rightarrow\mathbb{R}_+$ such that $\rho(x)>m(x)$ whenever $m(x)>0$. Define the relation $\psi$ from $X$ to $Y$ as $$ \psi(x)=\left\{\begin{matrix} \phi(x)\cap \{y\in Y:\|y\|<\rho(x)\} &\text{if}&\rho(x)>0\\ \{0\}&\text{if}&\rho(x)=0 \end{matrix}\right. $$ Then $\psi$ is lower hemicontinuous.

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Definition of hemicontinuity: A relation $\phi:X\rightarrow 2^Y$ is lower hemicontinuous if $\{x\in X: \phi(x)\cap V\neq\emptyset\}$ is open whenever $V\subset Y$ is open.

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Some rather obvious observations:

• $\phi(x)\neq\emptyset$ for all $x\in X$.
• One also can show directly from definition of lower hemicontinuity that for any open set $U\subset Y$ such that $\phi(x)\cap U\neq\emptyset$, the relation $\gamma(x)=\phi(x)\cap U$ is lower semicontinuous.

Is the last relation that the author mentions leads to the validity of the statement. It is not clear to me how to exploit that fact, and may attempts have not been successful thus far. At issue is that the open set $\{y\in Y:\|y\|<p(x)\}$ when $p(x)>0$ depends on $x$, and I have not been able to get some uniform control.

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Any hints or a sketch of a proof would be welcome.

Answer 1

Here is a sketch of a proof for the statement in the OP.

Notation. I will use l.h.c to denote lower hemicontinuous (of correspondences), and l.s.c to denote lower semicontinuous (of functions). I also use $B(y;r)$ to denote the open ball centered at $y$ with radius $r>0$.

It suffices to show that $\psi$ is lower hemicontinuous at every $x\in X$. Let $x\in X$.

First we make the observation that for any $x\in X$, from the definition of $m$, $p(x)>0$ implies that $$ \begin{align} \phi(x)\cap B(\mathbf{0};p(x))\neq\emptyset\tag{1}\label{one} \end{align} $$ for in this case, $0\leq m(x)<p(x)$.

1. Case $p(x)=0$. In this case $m(x)=0$, and since $\phi$ is closed valued, $\mathbf{0}\in\phi(x)$. Also $\psi(x)=\{\mathbf{0}\}$. Suppose $V\subset Y$ is open and $\mathbf{0}\in V$. Let $r>0$ such that $B(\mathbf{0};r)\subset V$. Since $\phi$ is l.h.c., there exists an open $W_x$ neighborhood of $x$ such that $$ \emptyset\neq \phi(z)\cap B(\mathbf{0};r)\qquad z\in W_x $$ For $z\in W_x$,

• If $p(z)=0$, then $\psi(z)\cap V=\{\mathbf{0}\}\neq\emptyset$.
• If $0<p(z)<r$, then by $\eqref{one}$ there exists $y\in \phi(z)\cap B(\mathbf{0};p(z))=\phi(z)\cap B(\mathbf{0};p(z))\cap V$.
• If $r\leq p(z)$, $\emptyset\neq \phi(z)\cap B(\mathbf{0};r)\subset\phi(z)\cap B(\mathbf{0};p(x))\cap V$.

2. Case $p(x)>0$. Suppose $V\subset Y$ is open and that $y_0\in\phi(x)\cap B(\mathbf{0};p(x))\cap V$. Then, there is $\delta>0$ such that $B(y_0;\delta)\subset V$. It follows that $\|y_0\|_Y+\delta<p(x)$, for $y_0+ru\in B(y_0;r)$ fir any unitary vector $u\in Y$. Since $p$ is l.s.c., there exists an open neighborhood $W_x$ of $x$ such that $$p(z)>\|y_0\|_Y+r,\qquad z\in W_x.$$ Since $\phi$ is l.h.c. there is an open neighborhood $V_x\subset W_x$ of $x$ such that $$ \emptyset\neq\phi(z)]\cap B(y_0;r),\qquad z\in V_x $$ If $z\in V_x$, $B(y_0;r)\subset B(\mathbf{0};p(z))\cap V$ and so, $\emptyset\neq \phi(z)\cap B(\mathbf{0};p(z))\cap V$.

Putting things together, we have shown that if $\psi(x)\cap V\neq\emptyset$, then there exists an one neighborhood $W_x$ of $x$ such that $$ \psi(z)\cap V\neq\emptyset\qquad\text{whenever}\quad z\in W_x$$

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Comments:

I took a look at E. Michael's paper and it seems that one can also prove the statement as a consequence of his Proposition 2.5 which states that if $V$ is an open entourage for some uniform structure on $Y$, $\phi$ and $\psi$ are l.h.c. correspondences from $X$ to $Y$, and $\theta(x)=\phi(x)\cap V(\phi(x))\neq\emptyset$ for all $x\in X$, then $\theta$ is a l.h.c. correspondence from $X$ to $Y$. The proof does not seem complicated, but it does require some knowledge of uniformities which are not in much used nowadays.

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