Let $I^{n}\overset{_\mathrm{def}}{=}\left\{x=\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{n}|\;\;a_i\leq x_{i} \leq b_{i}\,;\; i=1, \ldots, n\right\}$ be an $n$-dimensional closed interval and $I$ a closed interval $[a, b] \subset \mathbb{R}$.
Proposition. Let $f \in C^0(I^n\times I,\mathbb{R})$. If for all $(x_1,\ldots,x_n)\in I^n$ there is a $t^\ast\in [a,b]$ such that $f(x_1,\ldots,x_n,t^\ast)=0$ then the function $$ m(x_1,\ldots,x_{n-1},x_{n})=\inf\{t\in [a,b] : f(x_1,\ldots,x_{n-1},x_{n},t)=0\} $$ is also continuous on the $I^n$.
Can anyone suggest a mathematical analysis textbook that contains proof of this Proposition?
I have already sought proof of this result in classic textbooks of mathematical analysis such as
- Real Mathematical Analysis by Charles Chapman Pugh.
- Principles of Mathematical Analysis by Walter Rudin
- Mathematical Analysis I by V. A. Zorich
but without success.
An example: Let $g(t)=1$ for $t\leq 1$ and $g(t)=2-t$ for $t>1$. Let $[a, b]=[-3, 3]$. Let $$ f(x_1, ...,x_n, t)=(x_1^2+x_2^2+...+x_n^2+t^2)g(t), $$ Thus if $(x_1, ...,x_n)\neq 0$ we have $m(x_1, ...,x_n)=2$, while $m(0, ... ,0)=0$.
In general one can only show $m$ is lower semicontinuous.
If one assumes $t^*$ is unique, one can prove the conclusion, but then the $\inf$ in the definition of $m$ looks funny.