# Continuity of multivariable function

1. Feb 13, 2006

### Benny

Hi, I'm having trouble with the following question. I would like some help with it.

Q. A function $f:A \subset R^n \to R^m$ is continuous if and only if its component functions $f_1 ,...,f_m :A \to R$ are continuous.

Firstly, is there a difference between $C \subset D$ and $C \subseteq D$? Anyway in this question I need to do both directions.

The definition I have of continuity is:

$f:A \subset R^n \to R^m$ is continuous at $\mathop {x_0 }\limits^ \to \in A$ if

$$\mathop {\lim }\limits_{\mathop x\limits^ \to \to \mathop {x_0 }\limits^ \to } f\left( {\mathop x\limits^ \to } \right) = f\left( {\mathop {x_0 }\limits^ \to } \right)$$

Alternatively: given $\varepsilon > 0$, there exists $\delta > 0$ such that

$$\left\| {f\left( {\mathop x\limits^ \to } \right) - f\left( {\mathop {x_0 }\limits^ \to } \right)} \right\| < \varepsilon$$

$$\forall x \in A$$ satisfying $\left\| {\mathop x\limits^ \to - \mathop {x_0 }\limits^ \to } \right\| < \delta$.

(yes, it does say x is an element of A where x is just the normal scalar variable - I would've thought that x would be a vector in R^n in this case)

I'm not really sure where I should start. If I was to begin with f being continuous then I could write an equation saying that the limit as an arbitrary n-vector (call it (x_1,..,x_n)) approaches a fixed n-vector (call it (b_1,...,b_n)), is equal to f applied to the fixed n-vector. To show one side of the implication I would need to deduce that

$$\mathop {\lim }\limits_{x_i \to b_i } f\left( {x_i } \right) = f\left( {b_i } \right)$$ for each i = 1,...,n.

It just seems so immediate that continuity of f automatically leads to continuity of its components and vice versa. Is this just a matter of writing down a few equations or is it more complicated? Any help would be good thanks.

Last edited: Feb 13, 2006
2. Feb 13, 2006

### benorin

Firstly, the notation, and, in particular the vector stuff: the usual conventions are $$\vec{x}:=\left( x_1,\ldots ,x_n\right) \in\mathbb{R} ^n$$ and $$x:=\left( x_1,\ldots ,x_n\right) \in\mathbb{R} ^n$$. Either way is used in various texts.

Q. A function $f:A \subset R^n \to R^m$ is continuous if and only if its component functions $f_1 ,...,f_m :A \to R$ are continuous.

So your proof is two-fold: i. Prove the "if" part (a.k.a. sufficiency); ii. Prove the "only if" part (a.k.a. necessity).

3. Feb 13, 2006

### benorin

There a difference between $C \subset D$ and $C \subseteq D$, depending on the author's/text's convention, one is inclusive that is allows C=D, and the other is strict and hence does not admit C=D.

4. Feb 13, 2006

### benorin

The plan

If the "if" part: Assume that the component functions $f_1 ,...,f_m :A \to R$ are continuous. Prove that $f:A \subset R^n \to R^m$ is then continuous.

If the "only if" part: Assume that $f:A \subset R^n \to R^m$ is continuous. Prove that the component functions $f_1 ,...,f_m :A \to R$ are then continuous.

5. Feb 13, 2006

### HallsofIvy

Staff Emeritus
One fact that might simplify your calculations:
In any Rn,
$$d(x,y)= \sqrt{(x_1-y_1)^2+ (x_2-y_2)^2+ ...+ (x_n-y_n)^2}$$
and
$$d(x,y)= max (|x_1-y_1|, |x_2-y_2|, ..., |x_n-y_n|)$$
are equivalent metrics. That is, they give exactly the same open sets, continuity, etc. The second one makes looking at individual components easy.

6. Feb 13, 2006

### Benny

Thanks for the suggestions.

I haven't done anything on metric spaces but I assume that the concepts referred to are fundamental ones. If that's the case then I guess I can just look up some definitions. I'm not sure if the question intended the use of those things but it doesn't really matter.

I'm just having trouble piecing together what exactly I need to show. As I said in my initial post, I know that one definition of continuity is to have the limit of f(x) as x -> a being equal to f(a). That seems to be the easier of the two (that one and the one involving epsilons) to use in this case. But I'm not sure what to do. I guess I could write:

$$f:A \subset R^n \to R^m$$

$$f\left( {x_1 ,...,x_n } \right) = \left( {f_1 ,...,f_m } \right)$$

where by definition of f

$$\mathop {\lim }\limits_{\left( {x_1 ,...,x_n } \right) \to \left( {b_1 ,...,b_n } \right)} f\left( {x_1 ,...,x_n } \right) = f\left( {b_1 ,...,b_n } \right) = \left( {c_1 ,...,c_m } \right)$$

What does that tell me though?

I'm having immense trouble seeing through the symbols and figuring out what I need to do.

Last edited: Feb 13, 2006
7. Feb 13, 2006

### matt grime

Bear in mind we're dealing with this horrible mess:

f(x_1,....,x_n) = (f_1(x_1,...x_n),....,f_m(x_1,...,x_n))

Right. What is the distance between two vectors, forgetting the f for a while?

The distance between u=(u_1,..,u_m) and v=(v_1,...v_m) is

$$\sqrt{(u_1-v_1)^2 + \ldots +(u_m-v_m)^2$$

Now the statement u converges to v in this distance sense is if and only if this big square root tends to zero, which is if and only if u_i converges to v_i for all i, right?

If v is f(x) and u is f(a sequence tending to x), then convergence of the whole is iff convergence of the components is iff....

8. Feb 13, 2006

### arildno

It might be easiest to tackle first the "if f is cont, then each $f_{m}$ is cont."-statement.

Try to show that the following inequality must hold:
For any m, we have: $$|f_{m}|\leq||f||$$
Try to construct a proof using this insight.

9. Feb 13, 2006

### Benny

Thanks for the help guys. I'll try again and see if I can get something out.