4

Let be $X\subset \mathbb{R^m}$, $K\subset \mathbb{R^n}$ compact, $f : X\times K \rightarrow \mathbb{R^p}$ continuous and $c\in \mathbb{R^p}$. Suppose that for every $x\in X$, there is a unique $y \in K$ such that $f(x,y)=c$. Prove that the $y$ depends continuously from $x$.

-My idea was to take a function such that: $\varphi : X\rightarrow K$ where $$\varphi (x) = y$$ then you should to prove that $\varphi$ is continuous.

Thank you for your help.

Ysaac
  • 109

2 Answers2

3

Let $\varphi:X\to K$ be the function that you mentioned above. Let $x\in X$ and $(x_n)_{n\in\mathbb N}$ be a sequence with $x_n\to x$. Let $y_n := \varphi(x_n)$. Assume that $y_n \not\to \varphi(x)$. Then, by compactness there exists a subsequence $(y_{n_j})_{j\in\mathbb N}$ with $y_{n_j}\to z \neq \varphi(x)$.

Since $f$ is continuous $\displaystyle c = \lim_{j\to\infty}f(x_{n_j},y_{n_j}) = f(\lim_{j\to\infty}x_{n_j},\lim_{j\to\infty}y_{n_j}) = f(x,z)$.

Hence $z = \varphi(x)$. Contradiction.

Lukas Betz
  • 4,506
  • 1
    Excuse me, but I don't understand why do you assume that $(x_n)$ converge to x? I don't have information about $X$ – Ysaac May 11 '15 at 00:21
  • Because i am showing sequential continuity which is equivalent to continuity in metric spaces. That should be a familiar concept. – Lukas Betz May 11 '15 at 00:24
3

Sketch: There's nothing to prove at isolated points of $X$. Let $x$ be an arbitrary limit point of $X$, let $(x_{k})$ be an arbitrary sequence in $X\setminus\{x\}$ converging to $x$, and put $y_{k} = \varphi(x_{k})$ (using your definition of $\varphi$).

You know $K$ is compact, $f(x_{k}, y_{k}) = c$ for all $k$, and $y_{k}$ is the unique element of $K$ satisfying $f(x_{k}, y_{k}) = c$. Use these facts to show successively that:

  1. $(y_{k})$ has a subsequence that converges in $K$.

  2. Every subsequence of $(y_{k})$ that converges in $K$ has the same limit.

  3. $(y_{k})$ itself converges in $K$.

Why does this finish the job?

Andrew D. Hwang
  • 78,195
  • Thank you sir, but you say the $y_k$ is the unique element of K after you say $f(x_k,y_k)=c$ for all $k$?. I'm confused :( – Ysaac May 11 '15 at 00:32
  • In other words, for each $k$, $f(x_{k}, y_{k}) = c$, and if $f(x_{k}, y) = c$, then $y = y_{k}$. :) – Andrew D. Hwang May 11 '15 at 09:30
  • 1
    So this sequence is constant, right? regards. – Ysaac May 11 '15 at 12:25
  • 1
    The sequence $(y_{k})$ is the set of values $\bigl(\varphi(x_{k})\bigr)$; generally, this sequence is not constant. (The "uniqueness" in my post referred to for each $k$ there is a unique $y{k}$, but as $k$ varies, $y{k}$ does, too.) Incidentally, have you made a sketch of the set-up of the problem? That will surely help you visualize and organize the proof. :) – Andrew D. Hwang May 11 '15 at 12:31