Brouwer fixed points exist on every nonempty closed box

Let $J=\{j:a_j<b_j\}$ and $m=\mathrm{\mid }J\mathrm{\mid }$. If $m=0$, $X$ is a singleton and its sole point is fixed. If $m>0$, deleting the fixed coordinates identifies $X$ with a box $X_J$ whose side lengths are all positive. The coordinatewise affine map $T:[-1,1{]}^m\to X_J$ is a homeomorphism.

For $u\in D^m\setminus \{0\}$ define $h(u)=(\mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{/}\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }})u$ and set $h(0)=0$. Then $\mathrm{\parallel }h(u){\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }u{\mathrm{\parallel }}_2\le 1$, so $h(D^m)\subseteq [-1,1{]}^m$. For $y\mathrm{\ne }0$ in the cube define $h^{-1}(y)=(\mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\mathrm{\parallel }y{\mathrm{\parallel }}_2)y$ and set $h^{-1}(0)=0$; its Euclidean norm is $\mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, and direct substitution shows the formulas are inverse. Both maps are continuous away from zero. At zero, $\mathrm{\parallel }h(u){\mathrm{\parallel }}_2\le \sqrt{m}\mathrm{\parallel }u{\mathrm{\parallel }}_2$ and $\mathrm{\parallel }{h}^{-1}(y){\mathrm{\parallel }}_2=\mathrm{\parallel }y{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }y{\mathrm{\parallel }}_2$, so both are continuous there. Thus $h$ is a homeomorphism between the ball and cube.

The map $g=(T\circ h{)}^{-1}\circ f_J\circ (T\circ h):D^m\to D^m$ is continuous, where $f_J$ is the restriction of $f$ to the non-fixed coordinates after identifying the fixed coordinates. By [F1], $g$ has a fixed point $u$; then $(T\circ h)(u)$ is fixed by $f_J$, and restoring the fixed coordinates gives a fixed point of $f$ in $X$.