A continuous self-map of a nonempty closed box has a fixed point

Let $J=\{j:a_j<b_j\}$ and $m=\mathrm{\mid }J\mathrm{\mid }$. If $m=0$, $X$ is a singleton, so its sole point is fixed. If $m>0$, deleting the fixed coordinates identifies $X$ with a box $X_J$ whose side lengths are positive. The coordinatewise affine map $T:[-1,1{]}^m\to X_J$, with $T(u{)}_j=(a_j+b_j)\mathrm{/}2+(b_j-a_j)u_j\mathrm{/}2$, is a homeomorphism. Reinserting the fixed coordinate values defines $\iota :X_J\to X$, and projection onto the free coordinates defines ${\pi }_J:X\to X_J$; since the fixed output coordinates of $f$ equal their singleton values, $f_J={\pi }_J\circ f\circ \iota$ is a continuous self-map of $X_J$.

Define $\mathrm{\parallel }u{\mathrm{\parallel }}_2=({\sum }_j\mathrm{\mid }{u}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}={\max }_j\mathrm{\mid }{u}_j\mathrm{\mid }$, let $D^m=\{u:\mathrm{\parallel }u{\mathrm{\parallel }}_2\le 1\}$, and put $C_m=[-1,1{]}^m$. For $u\mathrm{\ne }0$ set $h(u)=(\mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{/}\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }})u$, and set $h(0)=0$. Then $\mathrm{\parallel }h(u){\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }u{\mathrm{\parallel }}_2\le 1$. For $y\mathrm{\ne }0$ in $C_m$, set $h^{-1}(y)=(\mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\mathrm{\parallel }y{\mathrm{\parallel }}_2)y$, and set $h^{-1}(0)=0$. For $u\mathrm{\ne }0$, $\mathrm{\parallel }h(u){\mathrm{\parallel }}_2=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2\mathrm{/}\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{\parallel }h(u){\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }u{\mathrm{\parallel }}_2$, so substituting in $h^{-1}$ gives $h^{-1}(h(u))=u$; the same norm calculation gives $h(h^{-1}(y))=y$ for $y\mathrm{\ne }0$. Both maps are continuous away from zero. For any vector $v$, one coordinate attains $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}^2\le {\sum }_j\mathrm{\mid }{v}_j{\mathrm{\mid }}^2$; each of the $m$ summands is at most $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}^2$, so $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\le m\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}^2$. Taking nonnegative square roots gives $\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }v{\mathrm{\parallel }}_2\le \sqrt{m}\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}$. Therefore for nonzero $u$, $\mathrm{\parallel }h(u){\mathrm{\parallel }}_2=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2\mathrm{/}\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}\le \sqrt{m}\mathrm{\parallel }u{\mathrm{\parallel }}_2$, and for nonzero $y$, $\mathrm{\parallel }{h}^{-1}(y){\mathrm{\parallel }}_2=\mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }y{\mathrm{\parallel }}_2$. These inequalities also hold at zero and show both maps are continuous there. Thus $h$ is a homeomorphism $D^m\to C_m$.

For $m>0$, $T\circ h$ is a homeomorphism from $D^m$ to $X_J$. The conjugate map $g=(T\circ h{)}^{-1}\circ f_J\circ (T\circ h):D^m\to D^m$ is continuous. By [F1], $g$ has a fixed point; mapping it back gives a fixed point of $f_J$, and applying $\iota$ gives a fixed point of $f$ in $X$.