Uniform continuity with respect to one variable?

Question

I want to prove the following result:

Let $K,$ $X,$ and $Y$ be metric spaces, and assume $K$ is compact. Then, if $F \colon K \times X \to Y$ is a continuous function and $x_0 \in X,$ for each $\varepsilon > 0$ there exists $\delta > 0$ such that whenever $d_X(x,x_0) < \delta,$ $d_Y(F(t,x), F(t,x_0)) < \varepsilon $ for every $t \in K.$

I guess this has something to do with the fact that any continuous function on a compact metric space is uniformly continuous, but I haven't been able to prove it.

Any help will be appreciated.

Answer 1

Let $\epsilon>0$ and fix $x_0\in X.$ For each $(t,x_0)\in K\times X,$ there is a $\delta_t$ (depending on $t$) such that if $(s,x)\in B_K(t,\delta_t)\times B_X(x_0,\delta_t)$, then $d(F(s,x), F(t,x_0))<\epsilon/2.$

But $K$ is compact, and $\{B_K(t,\delta_t):t\in K\}$ is an open cover of $K$, so we may pass to a finite subcover $\{B_K(t_i,\delta_i):1\le i\le n\}.$ Let $\delta=\min \{\delta_i:1\le i\le n\}$.

Now, suppose $d_X(x,x_0)<\delta.$ Of course, all this means is that $x\in B_X(x_0,\delta).$ Let $t\in K$ be arbitrary. Then, $t\in B_K(t_i,\delta_i)$ for some integer $1\le i\le n$ and, because $\delta<\delta_i,$ it follows that $(t,x_0)$ and $(t,x)$ are contained in $B_K(t_i,\delta_i)\times B_X(x_0,\delta_i)$, and therefore $d(F(t,x),F(t_i,x_0))<\epsilon/2$ and $d(F(t,x),F(t_i,x_0))<\epsilon/2.$ It now follows that

$d(F(t,x),F(t,x_0))\le d(F(t,x),F(t_i,x_0))+d(F(t_i,x_0),F(t_i,x))<\epsilon$, as desired.

Answer 2

Yes you are right. If you fix an $x_0\in X$ then the function $G:K\to Y$ with $G(x):=F(x,x_0)$ is uniformly continuous$.

Lets prove it. Let $\epsilon >0 $ then, as $G$ is continous for each $z\in K$ there is a $\delta_z >0$ such that for all $x\in K$ with $d_X(z,x)<\delta_z$ we have $d_Y(G(z),G(x)) \lt \epsilon/3$. Let us denote $B_z:=\{x\in K : d_X(z,x)<\delta_z \}$. Clearly $z\in B_z$. So if we consider the class of all $B_z$, then this is an open coverage of $K$. As $K$ is compact, there are $z_1, ..., z_n$ with $K=\cup_{i=1}^n B_{z_i}$. Set $$\delta = \min_{i=1,..,n}\{\delta_{z_i}/3\} >0$$

Now take a pair $a,b \in K$ with $d_X(a,b) \lt \delta $. Now $a$ has to be included in some $B_{z_k}$ and $b$ in $B_{z_j}$. Without loss of generality let us assume $\delta_{z_k} \le \delta_{z_j} $.

Then $$d_X(z_k,z_j) \le d_X(z_k,a)+d_X(a,b)+d_X(b,z_j)\lt \delta_{z_k}/3 + \delta + \delta_{z_j}/3 \le 3 \delta_{z_j}/3 =\delta_{z_j} $$ This tells us that $z_k \in B_{z_j}$ ! And therefore $d_Y(G(z_k),G(z_j))\le \epsilon/3$.

Now as $a\in B_{z_k}$, we have $d_Y(G(a),G(z_k)) \le \epsilon/3$ and for $b$ it similarly holds $d_Y(G(b),G(z_j)) \le \epsilon/3$

So in the end we have $$d_Y(G(a),G(b)) \le d_Y(G(a),G(z_k))+ d_Y(G(z_k),G(z_j)) + d_Y(G(z_j),G(b)) \lt \epsilon/3 + \epsilon/3 +\epsilon/3 =\epsilon$$

So $G$ is uniformly continuous.