# Relation between the norms on X and Y and the induced product norm on X x Y

Let $(X,||\cdot||_X)$, $(Y,||\cdot||_Y)$ be a pair of normed linear spaces, and $(X \times Y, ||\cdot||_{X \times Y})$ the induced product space and norm.

If $(x,y)$ is an element in $X \times Y$, is it true that $||(x,y)||_{X \times Y} \lt \delta$ implies that $||x||_{X \times Y} \lt \delta$, where $(||x||_{X\times Y}=||(x,0)||_{X\times Y}$)?

This has been true with every product norm that I have encountered, but I am not sure if it is true in general and/or if there is an easy way to show it.

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$\|x\|_{X\times Y}$ means $\|(x,0)\|_{X\times Y}$? –  a.r. Oct 10 '10 at 6:57
If you are defining the norm on the product by $||(x,y)||_{X\times Y}=||x||_X+||y||_Y$, then obviously the answer is "yes". How are you defining "the" norm in the product? –  Arturo Magidin Oct 10 '10 at 6:57
@Agusti: Yup. @Arturo: I'm not explicitly defining the norm. I'm wondering if the statement is true for every possible norm that you can define using the norm on the individual spaces. –  user1736 Oct 10 '10 at 14:55
@User1736: What does it mean to have a norm "defined using the norms on the individual spaces"? Perhaps you mean that the natural embeddings $X\to X\times Y$ and $Y\to X\times Y$ (given by $x\mapsto (x,0)$ and $y\mapsto (0,y)$) are continuous? –  Arturo Magidin Oct 10 '10 at 19:24
@User1736: Or perhaps that the projections $X\times Y\to X$ and $X\times Y\to Y$ are continuous? –  Arturo Magidin Oct 10 '10 at 19:59

No, this need not be true, unless I am missing some assumption implicit in your use of the phrase "the induced norm" (emphasis added). Here is an example where $X=Y=\mathbb{R}$. All norms on $X\times Y=\mathbb{R}^2$ are equivalent. The norm $\|(x,y)\|^2=2x^2-2xy+y^2=x^2+(x-y)^2$ does not satisfy your property because, for example $\|(x,x)\|=|x|$ while $\|(x,0)\|=\sqrt{2}|x|$. (This norm actually comes from an inner product on $\mathbb{R}^2$.)

The following was a result of misreading the question. I'll leave it here, for now at least, if for no other reason than leaving Arturo's correction comprehensible:

What is true is that the projection map $(x,y)\mapsto x$ is continuous. This implies as a special case the following (which turns out to actually be equivalent to continuity):

For all $\epsilon>0$, there is a $\delta>0$ such that for all $x\in X$ and $y\in Y$, $\|(x,y)\|<\delta$ implies $\|x\|<\epsilon$.

In fact, the projection is Lipschitz continuous, and $\delta$ can be taken to be $\epsilon$ divided by the Lipschitz constant. For the reason Qiaochu Yuan gave in a comment on Ross Millikan's answer, this is the best you could hope for.

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It would see, based on his comment response to Agusti Roig, that user1736 means $||x||$ to stand for $||(x,0)||$. That is, will $||(x,y)||\lt\delta$ imply $||(x,0)||\lt \delta$? (It's a strange question; my interpretation would have been yours and Qiaochu's, as well). –  Arturo Magidin Oct 10 '10 at 19:52
Oops, I didn't read that carefully! This answer isn't an answer at all. Thanks for catching my error. –  Jonas Meyer Oct 10 '10 at 19:54
It's not an error; the OP has been less than crystal clear on what he means... (-: –  Arturo Magidin Oct 10 '10 at 19:58
I've now corrected it, I think. –  Jonas Meyer Oct 10 '10 at 20:08
@User1736: the underscore confuses the renderer (it is also the signal for italics), as does the less-than symbol. For the latter, use \lt; for the former, you can precede it by a backslash to prevent it from messing up the renderer. –  Arturo Magidin Oct 10 '10 at 6:54
In fact, you can 'combine' the two norms and $X$ and $Y$ using any norm on $\mathbb R^2$, for example lp-norms. For $p=1$, you obtain you original product norm. –  Martin Dec 3 '10 at 11:43