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1) Can it happen that $\|x+a\|=\|x-a\|=\|x\|+\|a\|$ when $a\ne0$?

2) How large can $\min(\|x+a\|,\|x-a\|)/\|x\|$ be when $\|x\|\ge \|a\|$?

(For a inner-product space, the answers are no and $\sqrt{2}$ I think.)

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I changed \mathrm{min} to \min. That automatically provides proper spacing in expressions like $a\min b$ and in a displayed, as opposed to inline, setting, it affects the positions of subscripts, thus: $\displaystyle\min_{x\in A}$ ${}\qquad{}$ – Michael Hardy Jun 15 '13 at 18:01
I also changed numerous instances of $||x||$ to $\|x\|$. – Michael Hardy Jun 15 '13 at 18:02
up vote 5 down vote accepted

1) This is trivially true if $x=0$ or $a=0$. So you want $x\neq 0$ and $a\neq 0$. So what you are asking is equivalent to $$ \exists?\; x\neq 0, y\neq 0\qquad\Big\| \frac{x+y}{2}\Big\|=\Big\| \frac{x-y}{2}\Big\|=\frac{\|x\|+\|y\|}{2}. $$

Example: take $x=(1,1)$ and $y=(1,-1)$ in $\mathbb{R}^2$ with the $\ell^\infty$ norm $\|(x_1,x_2)\|=\max\{|x_1|,|x_2|\}$.

Inner-product space: recall the parallelogram law $\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$ which characterizes norms coming from an inner product space. So if $x,y$ satisfy your condition, we get $2(\|x\|+\|y\|)^2=2\|x\|^2+2\|y\|^2$ whence $\|x\|\|y\|=0$, i.e. $x=0$ or $y=0$. So the answer is indeed no.

Minkowski: more generally, if $1<p<\infty$, this is impossible in a $L^p$ space. Indeed, by the equality case of Minkowski inequality, if $f\neq 0$, $\|f+g\|=\|f\|+\|g\|$ forces $f=t g$ for $t>0$, and $\|f-g\|=\|f\|+\|g\|$ yields $f=-sg$ for $s> 0$. Thus $(s+t)g=0$ yields $g=0$.

2) For your second question, it suffices to restrict to the span of $x$ and $y$. Draw a picture of the two parallelograms one can build from $x$ and $y$.

Inner-product space: without loss of generality, $(x,y)\geq 0$ and $\|x-y\|$ is the minimum. So for $\|x\|$ and $\|y\|$ given, the maximum is reached for $(x,y)=0$ (in dimension $\geq 2$), in which case $$ \frac{\|x-y\|}{\|x\|}=\sqrt{\frac{\|x-y\|^2}{\|x\|^2}}=\sqrt{\frac{\|x-y\|^2}{\|x\|^2}}=\sqrt{\frac{\|x\|^2+\|y\|^2}{\|x\|^2}}=\sqrt{1+\frac{\|y\|^2}{\|x\|^2}}. $$ So under the constraint $\|x\|\geq \|y\|$, the maximum of the quotient is indeed $\sqrt{2}$.

General normed vector space: one can see by triangular inequality that $$ \frac{\min\{\|x-y\|,\|x+y\|\}}{\|x\|}\leq 2 \qquad \forall \|x\|\geq \|y\|>0. $$ The first example I gave shows that $2$ can be attained.

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For an inner product space the answer is no. Indeed:

$$\|x+a\|^2 = (x+a,x+a) = \|x\|^2 + \|a\|^2 + 2(x,a)\stackrel{!}{=} (\|x\| + \|a\|)^2$$ $$\Leftrightarrow 2(x,a) = 2\|x\|\cdot\|a\|$$

and this is true only if $x$ and $a$ are collinear. Now, for $x-a$ we find similarly:

$$-2(x,a) = 2\|x\|\cdot\|a\|$$

thus either $a$ of $x$ must be $0$.

For general vector spaces, the answer is yes. For example take the space $C[-1,1]$ of continuous functions on $[-1,1]$ with norm $\|f\| = \max_{x\in[0,1]}|f(x)|$ and consider the following two functions:

$$f(x) = 1,\ g(x) = x$$

Then $\|f+g\| = \|f-g\| = \|f\|+\|g\| = 2$.

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. . . and I did the same edit here as in the question. Even someone who routinely uses \stackrel misses this. Should this maybe be in an FAQ? – Michael Hardy Jun 15 '13 at 18:08
I think $\|f-g\|=\|1-x\|=1$ – Anonymous Coward Jun 15 '13 at 18:08
@dentisDark: Right, I meant to take $C[-1,1]$. I'll edit my answer. – Daniel Robert-Nicoud Jun 15 '13 at 18:38

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