Exercise 2.2.6

Question

Let $\mathbb{R}^{m}$ and $\mathbb{R}^{n}$ be Euclidean spaces. If $f: X \rightarrow \mathbb{R}^{m}$ and $g: X \rightarrow \mathbb{R}^{n}$ are continuous functions, show that $f \oplus g: X \rightarrow \mathbb{R}^{m+n}$ is also continuous. Is the converse true?

Prakash Anand · Accepted Answer

\begin{proof} Since $f, g$ are continuous, we know that for each $x_{0} \in X$, given $\epsilon^{\prime}>0$ there is $\delta_{1}, \delta_{2}$ such that if $d_{X}\left(x, x_{0} ight)<\delta_{1}$ then $d_{\left(\mathbb{R}^{m}, l^{2} ight)}\left(f(x), f\left(x_{0} ight) ight)<\epsilon^{\prime}$ and if $d_{X}\left(x, x_{0} ight)<\delta_{2}$ then $d_{\left(\mathbb{R}^{n}, l^{2} ight)}\left(g(x), g\left(x_{0} ight) ight)<\epsilon^{\prime}$.\par Let $x_{0} \in X$ and given $\epsilon>0$ let $\epsilon^{\prime}=\frac{\epsilon}{\sqrt{2}}$ and take $\delta=\min \left\{\delta_{1}, \delta_{2} ight\} .$ Now if $d_{X}\left(x, x_{0} ight)<\delta$ we know from the continuity of $f, g$ that $d_{\left(\mathbb{R}^{m}, l^{2} ight)}\left(f(x), f\left(x_{0} ight) ight)<\epsilon^{\prime}$ and $d_{\left(\mathbb{R}^{n}, l^{2} ight)}\left(g(x), g\left(x_{0} ight) ight)<\epsilon^{\prime}$.\par Note that $f(x), f\left(x_{0} ight)$ are $m$-tuples, and that $g(x), g\left(x_{0} ight)$ are $n$-tuples. So let $f_{i}(x)$ denote the $i$-th component in the $m$-tuple $f(x)$, and similarly for $f_{i}\left(x_{0} ight), g_{i}(x), g_{i}\left(x_{0} ight)$. And so we will define $f \oplus g(x)=\left(f_{1}(x), f_{2}(x), \ldots, f_{m}(x), g_{1}(x), g_{2}(x), \ldots, g_{n}(x) ight)$. So, $$ \begin{aligned} &d_{\left(\mathbb{R}^{m}, l^{2} ight)}\left(f(x), f\left(x_{0} ight) ight)<\epsilon^{\prime} ext { and } d_{\left(\mathbb{R}^{n}, l^{2} ight)}\left(g(x), g\left(x_{0} ight) ight)<\epsilon^{\prime} \ &\Leftrightarrow \sqrt{\sum_{i=1}^{m}\left(f_{i}(x)-f_{i}\left(x_{0} ight) ight)^{2}<\epsilon^{\prime} ext { and } \sqrt{\sum_{i=1}^{n}\left(g_{i}(x)-g_{i}\left(x_{0} ight) ight)^{2}<\epsilon^{\prime}}} \ &\Rightarrow \sum_{i=1}^{m}\left(f_{i}(x)-f_{i}\left(x_{0} ight) ight)^{2}<\left(\epsilon^{\prime} ight)^{2} ext { and } \sum_{i=1}^{n}\left(g_{i}(x)-g_{i}\left(x_{0} ight) ight)^{2}<\left(\epsilon^{\prime} ight)^{2} \ &\Rightarrow \sum_{i=1}^{m}\left(f_{i}(x)-f_{i}\left(x_{0} ight) ight)^{2}+\sum_{i=1}^{n}\left(g_{i}(x)-g_{i}\left(x_{0} ight) ight)^{2}<2\left(\epsilon^{\prime} ight)^{2} \ &\Rightarrow \sqrt{\sum_{i=1}^{m}\left(f_{i}(x)-f_{i}\left(x_{0} ight) ight)^{2}+\sum_{i=1}^{n}\left(g_{i}(x)-g_{i}\left(x_{0} ight) ight)^{2}}<\sqrt{2} \epsilon^{\prime}=\epsilon \end{aligned} $$ Thus, $d_{\left(\mathbb{R}^{m+n}, l^{2} ight)}\left(f \oplus g(x), f \oplus g\left(x_{0} ight) ight)<\epsilon$. Therefore $f \oplus g$ is continuous. \end{proof} The converse is true, and the proof is basically the same, but we will give it anyway. \begin{proof} If $f \oplus g$ is continuous then we know for each $x_{0} \in X$ given $\epsilon^{\prime}>0$ there is $\delta^{\prime}>0$ such that if $d_{X}\left(x, x_{0} ight)<\delta^{\prime}$, then $d_{\left(\mathbb{R}^{2}, l^{2} ight)}\left(f \oplus g(x), f \oplus g\left(x_{0} ight) ight)<\epsilon^{\prime}$. Let $x_{0} \in X$, and given $\epsilon>0$ let $\epsilon^{\prime}=\epsilon$ and take $\delta^{\prime}=\delta$. Now from the continuity of $f \oplus g$ we have: $$ \begin{aligned} &\Rightarrow \sqrt{\sum_{i=1}^{m}\left(f_{i}(x)-f_{i}\left(x_{0} ight) ight)^{2}+\sum_{i=1}^{n}\left(g_{i}(x)-g_{i}\left(x_{0} ight) ight)^{2}}<\epsilon^{\prime}=\epsilon \ &\Rightarrow \sum_{i=1}^{m}\left(f_{i}(x)-f_{i}\left(x_{0} ight) ight)^{2}+\sum_{i=1}^{n}\left(g_{i}(x)-g_{i}\left(x_{0} ight) ight)^{2}<\epsilon^{2} \ &\Rightarrow \sum_{i=1}^{m}\left(f_{i}(x)-f_{i}\left(x_{0} ight) ight)^{2}<\epsilon^{2} ext { and } \sum_{i=1}^{n}\left(g_{i}(x)-g_{i}\left(x_{0} ight) ight)^{2}<\epsilon^{2} \end{aligned} $$ $$ \Leftrightarrow \sqrt{\sum_{i=1}^{m}\left(f_{i}(x)-f_{i}\left(x_{0} ight) ight)^{2}}<\epsilon ext { and } \sqrt{\sum_{i=1}^{n}\left(g_{i}(x)-g_{i}\left(x_{0} ight) ight)^{2}}<\epsilon $$ Hence, $d_{\left(\mathbb{R}^{m}, l^{2} ight)}\left(f(x), f\left(x_{0} ight) ight)<\epsilon$ and $d_{\left(\mathbb{R}^{n}, l^{2} ight)}\left(g(x), g\left(x_{0} ight) ight)<\epsilon$ Thus $f, g$ are also continuous. \end{proof} \begin{remark} Note that there are some other ways to do this exercise: One way is to use Lemma 12.1.18 $(a \Leftrightarrow d)$, and note that convergence in $d\left(\mathbb{R}^{m}, l^{2} ight), d\left(\mathbb{R}^{n}, l^{2} ight)$ implies convergence of each component, which implies convergence in $d\left(\mathbb{R}^{m+n}, l^{2} ight) .$ \end{remark} \begin{remark} Also one could use the same approach as my proof, but note that: $$ d_{\mathbb{R}^{m+n}, l^{2}}^{2}\left(f \oplus g(x), f \oplus g\left(x_{0} ight) ight)=d_{\mathbb{R}^{m}, l^{2}}^{2}\left(f(x), f\left(x_{0} ight) ight)+d_{\mathbb{R}^{n}, l^{2}}^{2}\left(g(x), g\left(x_{0} ight) ight) $$ \end{remark}

Exercise 2.2.6

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Exercise 2.2.6

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