# Question about direct sum of function space

On page 30 about direct sums on vector spaces it says:

Let $V$ be the space of continuous functions on a domain $U \subset \mathbb R^3$. Then $V \oplus V \oplus V$ is the space of continuous $\mathbb R^3$-valued functions on $U$.

What I don't understand is how this threefold direct sum of $V$ can lead to a statement about the codomain ("-valued" functions). But I just guess that I don't get it altogether - please enlighten me. Thank you!

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I think what confuses me most is that we start with $\mathbb R^3$ and end up with it. It would come more natural if we started with $\mathbb R$ and after taking the direct sum three times would end in $\mathbb R^3$.

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A function to $\mathbb R^3$ is continuous iff its component scalar valued functions are continuous – Dactyl Jun 9 '11 at 16:41
@Dactyl: Could you please elaborate? So I don't get it... – vonjd Jun 9 '11 at 16:46
The first instance of $R^3$ is the domain and $R$ is the co-domain, in the second instance, both are $R^3$. – Dactyl Jun 9 '11 at 18:15

What is unclear in the exposition is just exactly what the author means by "the space of continuous functions on a domain $U\subseteq \mathbb{R}^3$."

What he means is that $U$ is the vector space of real-valued continuous functions with domain $U\subseteq\mathbb{R}^3$; that is, the elements of $U$ are function $f\colon U\to\mathbb{R}$ that are continuous, with $U\subseteq \mathbb{R}^3$. (He omits "real valued" when he says "functions").

That means that an element of $V\oplus V\oplus V$ consists of a $3$-tuple of real valued functions, $(f_1,f_2,f_3)$, with $f_i\colon U\to\mathbb{R}$. Thus, the tuple is naturally interpreted as a function from $U$ to $\mathbb{R}^3$ in which each component is continuous.

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The space $V\oplus V \oplus V$ is nothing else than the set of triples $$(v_1,v_2,v_3)$$ where $$v_1,v_2,v_3\in V.$$ So the set $U$ is being mapped to 3-component objects which I will call "vectors". Each element $P$ of $U$ is assigned a 3-component vector $(v_1(P), v_2(P), v_3(P))$. The continuity of a vector-valued function, as Dactyl pointed out, means that each of the components of the vector is a continuous function. Indeed, it means that each $v_i$ among the three components is an element of $V$.

One may either say that it is a definition of the continuity of vector-valued functions.

Equivalently, one may define the continuity in this case by looking at balls in the space ${\mathbb R}^3$ into which we're mapping things. But in the manipulation with $\epsilon$ and $\delta$, balls are equivalent to little cubes - one may squeeze a ball inside a cube or vice versa (at least for finite dimensions such as 3), so we may equivalently replace the balls by the cubes, and then the definition of continuity using balls gets reduced to that of the cubes, i.e. to the component-wise continuity.

At any rate, the continuity of 3-component functions is either defined as - or may be proved to be - the continuity of each component.

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I think what confuses me is that we start with $\mathbb R^3$ and end up with it. It would come more natural if we started with $\mathbb R$ and after taking the direct sum three times would end in $\mathbb R^3$. – vonjd Jun 9 '11 at 17:11
Dear @vonjd, your problem already starts with ${\mathbb R}^3$: the first sentence says that $V$ itself are functions from $U$ which is a subset of ${\mathbb R}^3$. $V$ itself are functions that take value in ${\mathbb R}$ and $V\oplus V\oplus V$ are functions from ${\mathbb R}^3$ to another ${\mathbb R}^3$. But functions from ${\mathbb R}$ never appear in your problem at all so I don't understand in what sense it would be "natural". Real numbers and their 3rd power are equally natural but only the latter appear in your problem. The tripling only affects the value of the function not the domain – Luboš Motl Jun 10 '11 at 5:23

In fact, the domain $U$ does not play an actual role here, specifying that it is a subset of $\mathbb R^3$ is just adding to the confusion, in my opinion.

First let us establish that $V$ is indeed a vector space. $f\in V$ if and only if $f\colon U\to\mathbb R$ and $f$ is continuous.

If $f\in V$ and $\alpha\in\mathbb R$ then $g(x) = \alpha\cdot f(x)$ is indeed a continuous function whose domain is $U$ and codomain $\mathbb R$, and if $f,g\in V$ then the function $h(x) = f(x) + g(x)$ is also a continuous function whose domain is $U$ and codomain is $\mathbb R$.

Now, the definition of $V\oplus V\oplus V$ is simply $\{(f,g,h)\mid f,g,h\in V\}$.

Suppose $(f,g,h)\in V\oplus V\oplus V$, then it defines a function: $h(x) = ( f(x), g(x), h(x))$. The function $h$ is continuous if and only if it is continuous in every coordination, and since $f,g,h$ are all continuous then so is $h$.

Since the domain of $f,g,h$ is $U$, then $h\colon U\to\mathbb R^3$, as it takes a point in $U$ and returns a vector in $\mathbb R^3$.

On the other hand, if $h$ is a continuous function from $U$ to $\mathbb R^3$, then $h(x) = (h_1(x), h_2(x), h_3(x))$, and $h_i$ are continuous functions from $U$ to $\mathbb R$, so $h_1,h_2,h_3\in V$ as needed.

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I guess you can identify the space of continuous functions on $U$ that take values in $\mathbb{R}$ with subspaces of the space of continuous functions on $U$ that take values in $\mathbb{R}^3$ by isomorphisms of type $x\mapsto (x,0,0)$, $x\mapsto (0,x,0)$, $x\mapsto (0,0,x)$. In this view the book's assertion may look more natural, because now you're indeed having the direct sum of subspaces of a given space.

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I think what confuses me is that we start with $\mathbb R^3$ and end up with it. It would come more natural if we started with $\mathbb R$ and after taking the direct sum three times would end in $\mathbb R^3$. – vonjd Jun 9 '11 at 17:12
@vonjd: it all depends on how the author talks about direct sum. When I took LA, direct sums where defined for subspaces of a given linear space. But generally in algebra you can talk about direct sums as structures defined on $V_1\times ... \times V_n$, where the operation(s) is defined as is most natural (e.g., $(v_1,...,v_n)+(w_1,...,w_n)=(v_1+w_1,...,v_n+w_n)$). But in the end, these approaches are pretty much the same. – Weltschmerz Jun 9 '11 at 17:16