Show that the function $h : A \to F \times G$ defined by $h(x) = (f(x), g(x))$ is uniformly continuous.

Question

Let $(E,\|\cdot\|_E)$, $(F,\|\cdot\|_f)$, $(G,\|\cdot\|_G)$ three normed vector spaces. We endow the space $F \times G$ with the norm $\|(y,z)\|= \|y\|_F + \|z\|_G$. Let $A \subset E$ and let $f : A \to F$ and $g: A \to G$ be two uniformly continuous function. Show that the function $h : A \to F \times G$ defined by $h(x) = (f(x), g(x))$ is uniformly continuous.

I'm blocked on this problem. Could anyone give me a hint to solve the problem?

Answer 1

Remember that, by definition, we must prove that $ \forall \epsilon > 0 , \exists \ \delta > 0 $ so that if $ ||x-y||_E < \delta $ then $ ||h(x) - h(y) ||<\epsilon $

We have, above, chosen an arbitrary $ \epsilon$ and, by hipotesys, we have $ \delta_f > 0$ such that $||f(x)-f(y)||< \epsilon/2 $ and similarly we have $ \ \delta_g >0 $ such that $ ||g(x)-g(y)||< \epsilon/2 \ \ $ IF $x \ and\ y\ satisfy\ ||x-y||_E<\delta_F \ \ AND\ \ ||x-y||_E<\delta_G \ $ at the same time.

So if I take $\delta$ as the minimum of both deltas, I can be sure that if $||x-y||_E<\delta $ then both$||f(x)-f(y)||< \epsilon/2 $ and $ ||g(x)-g(y)||< \epsilon/2 \ \ $ . Let's see that this $\delta$ is the one we are looking for:

If $||x-y||_E<\delta $ then $$||h(x)-h(y)||=||(f(x),g(x))-(f(y),g(y))||=||(f(x)-f(y),g(x)-g(y))|| = \\ ||f(x)-f(y)||_F + ||g(x)-g(y)|_G \leq \epsilon/2 + \epsilon/2= \epsilon $$

Where I used the observation of the preceding paragraph only at the inequality, the rest was definition of your function.

Answer 2

Do you know the definition of "uniformly continuous"? Just follow the definition! Let $\epsilon >0$ . We look for a $\delta>0 / ||h(x)-h(y)||< \epsilon \ \ if \ ||x-y||_E < \delta $.

You know that $\exists \ \delta_f , \ \delta_g >0 / ||f(x)-f(y)||< \epsilon/2 \ \ $ and similarly $ ||g(x)-g(y)||< \epsilon/2 \ \ $(by hypotesis) ....etc.