1
$\begingroup$

I have a statement I cannot figure out whether it's true or not. Let $(X,d_X)$, $(Y,d_Y)$ and $(Z,d_Z)$ be compact metric spaces, and $f\colon X\times Y\to Z$ be continuous. Let $(x_n)_n\in X^\mathbb N$ be a sequence in $X$ such that $x_n\to x$. I am trying to figure out whether the sequence of functions $h_n\colon Y\to Z$ defined by \begin{equation} h_n(y)=f(x_n,y) \end{equation} converges uniformly to $h(y)=f(x,y)$. Pointwise convergence is clear, and I know that $f$ is uniformly continuous. The problem I can't decide on, specifically, is whether the "$N$" in the statement for convergence ($\forall \epsilon>0\; \exists N\in\mathbb N$ s.t. ...). of the $h_n$ functions depends on both $\epsilon$ and $y$, or can be made independent of the latter.

$\endgroup$
2
  • $\begingroup$ Which metric on $X \times Y$ do you use? $\endgroup$
    – enedil
    Jun 5, 2018 at 22:12
  • $\begingroup$ Forgot to mention: $d((x,y),(x',y'))=d_X(x,x')+d_Y(y,y')$ $\endgroup$
    – Enrico
    Jun 5, 2018 at 22:17

1 Answer 1

1
$\begingroup$

Yes, it is true. Take $\varepsilon>0$. Now, take $\delta>0$ such that$$d_X(x_1,x_2)<\delta\wedge d_Y(y_1,y_2)<\delta\implies d_Z\bigl(f(x_1,y_1),f(x_2,y_2)\bigr)<\varepsilon.$$Now, if $p\in\mathbb N$ is such that $n\geqslant p\implies d_X(x_n,x)<\delta$, then\begin{align}n\geqslant p&\implies d_X(x_n,x)<\delta\wedge d_Y(y,y)=0<\delta\\&\implies d_Z\bigl(f(x_n,y),f(x,y)\bigr)<\varepsilon\end{align}

$\endgroup$
2
  • $\begingroup$ Wait, why in the fourth line you have $n\geq p \implies d_X(x_n,x)<\delta$? Can't it happen that $\epsilon\geq \delta$? I'm a bit lost there. $\endgroup$
    – Enrico
    Jun 5, 2018 at 22:34
  • $\begingroup$ @Enrico My mistake was on the third line; I wrote $\varepsilon$ instead of $\delta$. I suppose that everything is well now. $\endgroup$ Jun 5, 2018 at 22:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .