Continuity of vector space operations in a normed space

Question

Here's problem 4 immediately following section 2.3 in Erwine Kryszeg's book, Introductory Functional Analysis With Applications:

Show that in a normed space $X$, vector addition and scalar multiplication are continuous operations with respect to the norm; that is, the mappings defined by $(x,y) \mapsto x+y$ and $(\alpha,x) \mapsto \alpha x$ are continuous.

Now the map $(x,y) \mapsto x+y$ is a map from $X \times X$ to $X$; so we can consider $X\times X$ under the norm defined as follows: $$|| (x,y) ||_{X\times X} \colon= ||x||_X + ||y||_X$$ for all $(x,y) \in X \times X$. With this norm, we can easily prove the vector addition map to be continuous.

But what about the norm on $K \times X$ for the continuity of the scalar multiplication map? Here $K$ (either $\mathbb{R}$ or $\mathbb{C}$) denotes the field of scalars.

Answer 1

You can use the same idea that works for $+$: $$\|(\alpha,x)\|_{K\times X}=|\alpha|+\|x\|.$$ In any case, you can endow $K\times X$ with the product topology and no explicit norm is required.

EDIT:

$$\eqalign{\|\alpha x - \alpha_0 x_0\|_X & = \|\alpha x - \alpha_0 x + \alpha_0 x - \alpha_0 x_0\|_X\cr &\le\|\alpha x - \alpha_0 x\|_X + \|\alpha_0 x - \alpha_0 x_0\|_X\cr & = |\alpha - \alpha_0|\cdot\|x\|_X + |\alpha_0|\cdot\|x-x_0\|_X\cr &\le(1+\|x_0\|_X)\cdot|\alpha - \alpha_0|+ |\alpha_0|\cdot\|x-x_0\|_X\cr &\le(1+\|x_0\|_X+|\alpha_o|)\|(\alpha-\alpha_0,x-x_0)\|_{K\times X}\cr & = (1+\|x_0\|_X+|\alpha_o|)\|(\alpha,x)-(\alpha_0,x_0)\|_{K\times X}.}$$ (Why $\|x\|_X\le 1+\|x_0\|_X$?)

Answer 2

The fields $\mathbb R$ and $\mathbb C$ are equipped with a standard topology (derived from the standard metric and the standard absolute value).

Answer 3

There are many forms of product norm. See, for example, P153 of Folland "Real Analysis". Instead of the additive form of product norm $\|(x,y)\|=\|x\|+\|y\|$, here I use max form $\|(x,y)\|=\max(\|x\|,\|y\|)$ to prove continuity of scale multiplication $(\alpha,x)\mapsto \alpha x$. Note that the max form is equivalent to additive form, so they induce the same metric topology. Specifically, $\|(\alpha,x)\|_{K\times X}=\max(|\alpha|,\|x\|)$. From the definition of max form of product norm, it is easy to see $|\alpha|,\|x\|\le\max(|\alpha|,\|x\|)=\|(\alpha,x)\|_{K\times X}$. Denote the induced metric in $K\times X$ as $d$.

Now fix $\alpha_0$ and $x_0$. If $x_0=0$, $(\alpha,x)\mapsto \alpha x$ is a constant function, which is trivially continuous. In the following, we assume $x_0\ne0$.

Let $$\begin{equation*} A=\begin{cases} \alpha_0+1 &\text{if $\alpha_0\ge0$,}\\ -(\alpha_0-1) &\text{otherwise.} \end{cases} \end{equation*} $$ For any $\epsilon>0$, let $$\delta=\min(1,\frac{\epsilon}{2\|x_0\|},A,\frac{\epsilon}{2A}).$$ Then as long as $d\bigl((\alpha,x),(\alpha_0,x_0)\bigr)<\delta$, we have $|\alpha x-\alpha_0x_0|<\epsilon$, establishing the continuity of scale multiplication. This is because, when it is the case, $$\eqalign{d\bigl((\alpha,x),(\alpha_0,x_0)\bigr) &=\|(\alpha,x)-(\alpha_0,x_0)\|_{K\times X}\cr &=\|(\alpha-\alpha_0,x-x_0)\|_{K\times X}\cr &=\max(|\alpha-\alpha_0|,\|x-x_0\|)\cr &<\delta,}$$ so $|\alpha-\alpha_0|<1$ which forces $|\alpha|<A$. Likewise, we have $|\alpha-\alpha_0|<\frac{\epsilon}{2\|x_0\|}$ and $\|x-x_0\|<\frac{\epsilon}{2A}$. As a result, $$\eqalign{\|\alpha x - \alpha_0 x_0\| & = \|\alpha x - \alpha x_0 + \alpha x_0 - \alpha_0 x_0\|\cr &\le\|\alpha x - \alpha x_0\| + \|\alpha x_0 - \alpha_0 x_0\|\cr & = |\alpha|\|x-x_0\| + |\alpha-\alpha_0|\|x_0\|\cr &<A\frac{\epsilon}{2A} + \frac{\epsilon}{2\|x_0\|}\|x_0\|\cr & = \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.}$$