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I need to prove or refute that for every norm in $\mathbb{R}^n$:$ \left \| x \right \|\leq \max (\left \| x+y \right \|,\left \| x-y \right \|)$. It's been quite a while since I studied Linear algebra 1. I tried to look for vectors $x$ and $y$ such that they will refute the claim, but I didn't find any, so I tried to prove the question by showing the explicit sum of each norm, and go on form that, but it didn't do either.

(sorry if the question is too easy or silly)

Any help?

Thank you very much!

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up vote 12 down vote accepted

$2x = (x + y) + (x - y)$. So the triangle inequality gives $2\|x\| \le \|x+y\| + \|x-y\|$.

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A geometric way to think of this is to note that balls in normed spaces are convex, and $x$ lies on the line segment between $x+y$ and $x-y$. If $\|x\|$ were greater than $\|x-y\|$ and $\|x+y\|$, then $\{z:\|z\|<\|x\|\}$ would be an open ball that contains $x-y$ and $x+y$, but not their midpoint $x$, and therefore it would not be convex.

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The reason you found no counterexamples is that the claim is true. Suppose $\|x\|>\|x+y\|$ and $\|x\|>\|x-y\|$ for some $x,y\in \mathbb{R}^n$. You can project these down to $\mathbb{R}$ and $\mathbb{R}^{n-1}$ respectively to get a counterexample in at least one of these, so eventually you get a counterexample in $\mathbb{R}$. But in this case $x$ either has the same sign as or a different sign than $y$, and either way the inequality holds, contradicting the existence of a counterexample.

EDIT: As Jonas Meyer pointed out, this only works for inner product spaces where the norm is induced by the inner product. I have left it up so that those who've read it are aware of this failure.

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Is there a very easy way to see that you "get a counterexample in at least one of these"? – Jonas Meyer Jan 17 '12 at 9:00
@JonasMeyer Yes. If $\|(x_2,\ldots,x_n)\|\leq\|(x_2+y_2,\ldots,x_n+y_n)\|$ and $|x_1|\leq |x_1+y_1|$ then by orthogonality $\|(x_1,\ldots,x_n)\|\leq\|(x_1+y_1,\ldots,x_n+y_n)\|$ and similarly for $\|x-y\|$. – Alex Becker Jan 17 '12 at 9:09
What do mean by orthogonality? It is never stated that there is an inner product, let alone the standard inner product. – Jonas Meyer Jan 17 '12 at 9:10
@JonasMeyer You're quite right. My intuition got ahead of my brain at this hour. – Alex Becker Jan 17 '12 at 9:11
Regarding the edit, I'm not saying it isn't true for all norms. I at least partially convinced myself that it is true in general, but it didn't seem "very easy," which is why I asked. (And by "the" inner product, do you mean the standard inner product, a.k.a. the "dot product"?) – Jonas Meyer Jan 17 '12 at 9:17

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