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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
up vote 3 down vote accepted

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

Your question doesn't show up well, but the wikipedia entry Normed vector space claims that if you have the product of a finite number of normed vector spaces, the sum of the norms in the individual spaces is a norm in the product.

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Yeah, I just took out the tex formatting, so hopefully it's a little better now. I know that the sum of the norms is an example of a product norm, and the statement is certainly true for this norm. However, I'm wondering if the statement is true for any possible product norm induced from the norms on the individual spaces. – user1736 Oct 10 '10 at 4:50
@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
@user1736: this is surely false. Any scalar multiple of an induced norm is also an induced norm. – Qiaochu Yuan Oct 10 '10 at 10:46
@Arturo: Thanks for the tips. @Qiachu: Yeah, I think you're right. Thanks for your help. – user1736 Oct 10 '10 at 15:04
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. – shuhalo Dec 3 '10 at 11:43

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