Simple Triangle-like Inequality for a general norm

Question

I came up with this inequality as I was thinking about norms. I don't know if it's true though.

Given three points $x,y,z$ in a normed space $(X,\|\cdot\|)$, is it true that if

$$ \|x-y\|\le\|y\| $$

and

$$ \|x-z\|\le\|z\| $$

then

$$ \|x-(y+z)\|\le\|y+z\| ? $$

In other words, for a given vector $x$ in a normed space $X$, is the set $K:=\{y:\|x-y\|\le\|y\|\}$ additive?

First, if it were true, it almost looks like we could change the $y$ and $z$ around so that, e.g. the 1st and 3rd inequalities imply the 2nd. As it is, given any arrangement $x+y=z$ (none of which is $0$), we can tell which is which by simply adding.

I thought proving $X=\mathbb{R}$ would easily turn into $X=l_p$, but that seems naive.

Inner Product Space

The first two inequalities reduce to

$$ \langle x,x\rangle\le 2\langle y,x\rangle $$

and

$$ \langle x,x\rangle\le 2\langle z,x\rangle $$

Adding these together gets

$$ \|x\|^2\le \langle y+z,x\rangle\le 2\langle y+z,x\rangle $$

Working the same algebra backwards then gets the third inequality.

Other norms?

Note that if $y\in K$, then $cy\in K$ for all $c\ge 1$. So there is a curve/surface in $X$, symmetric about $0$ (AKA odd), that divides $X$ into two parts, one part being $K$. It looks like if the unit ball $B$ of $X$ is oblong and $x$ is not perpendicular to an axis of symmetry of B, the curve/surface will not be linear and hence not convex (symmetric about $0$ and convex implies linear).

So, the factor of $2$ cannot be dismissed for general $X$ as it was for inner product spaces (?). I feel like that argument says the 2 cannot be dismissed for some inner product spaces as well. Mistake?

Thanks!

Answer 1

The inequality can fail for $X=\Bbb R^2$ endowed with $\|\cdot\|_\infty$ norm. Indeed, let $x=(1,0)$, $y=(-1,2)$, and $z=(-1,-2)$. Then $\|x-y\|=\|y\|=\|x-z\|=\|z\|=2$, but $\|x-(y+z)\|=3>2=\|y+z\|$.

Answer 2

Update: Total rewrite! … And then partial retraction. Hopefully, some part of this is still helpful and I will continue to think about how to get a possibly complete answer. This is a great question. I've spent quite a while thinking about it now, as well as perusing Dan Amir's Characterizations of Inner Product Spaces for possibly related conditions. I thought I had a solution, but I made a sign error that you (Aaron Goldsmith) pointed out.

First, some definitions. Let $(V,||\cdot||)$ be a normed vector space. For any $x\in V$, define $$K_x=\big\{y\in V:||x-y||\leq ||y||\big\}\qquad\text{and}\qquad J_x=\big\{y\in V:||x-y||\leq ||x+y||\big\}.$$ For any $x\in V$, notice that $K_{2x}=x+J_x$ and thus $\partial K_{2x}=x+\partial J_x$. Note also that $$\partial J_x=\big\{y\in V:||x-y||=||x+y||\big\}=J_x\cap J_{-x},$$ for any $x\neq 0$. Now, I can state my partial result.

Proposition. Consider the following conditions:

1. $||\cdot||$ arises from an inner product on $V$;

2. $J_x$ is convex for all $x\in V$;

3. $\partial J_x$ is convex for all $x\in V$;

4. $J_x$ is additive for all $x\in V$;

5. $\partial J_x$ is additive for all $x\in V$;

6. $K_x$ is convex for all $x\in V$;

7. $\partial K_x$ is convex for all $x\in V$;

8. $K_x$ is additive for all $x\in V$.

I conjecture that $(8) \Longrightarrow (1)$, but have yet to prove it. We do know that $$(1)\Longleftrightarrow(2) \Longleftrightarrow (3) \Longleftrightarrow (4) \Longleftrightarrow (5) \Longleftrightarrow (6) \Longleftrightarrow (7) \Longrightarrow (8).$$

Proof. The equivalences $(1)\Longleftrightarrow (3)$ and $(1)\Longleftrightarrow(5)$ are conditions $(3.2)$ and $(4.13)$ in Amir's book, respectively. Since convexity is preserved by translation, we also have $(2)\Longleftrightarrow(6)$ and $(3)\Longleftrightarrow(7).$ To complete the proof, we will show that $(2)\Longrightarrow (3)$ and $(4)\Longrightarrow (5)$, and that $(1)\Longrightarrow (2),$ $(4)$, $(6)$ and $(8)$.

$(2)\Longrightarrow (3)$ and $(4)\Longrightarrow (5)$: Suppose that $J_x$ is convex (resp. additive) for all $x\in V$. First note that $\partial J_0=\partial V=\emptyset$ is vacuously convex and additive. If $x\neq 0$, then $\partial J_x=J_x\cap J_{-x}$ is convex (resp. additive) because convex (resp. additive) sets are closed under intersection.

$(1)\Longrightarrow (2)$, $(4)$, $(6)$ and $(8)$: Suppose that $||x||^2=\langle x,x\rangle$ for all $x\in V$, where $\langle \,\cdot\,,\cdot\,\rangle$ is an inner product on $V$. Then we have $$J_x=\big\{y\in V:\langle x,y\rangle \geq 0\big\}\qquad\text{and}\qquad K_x=\big\{y\in V:2\langle x,y\rangle \geq ||x||^2\big\}.$$ These sets are both convex and additive, since they are inverse images of sets that are convex and additive under the linear map $y\mapsto \langle x,y\rangle$. $\square$

The proof of $(8)\Longrightarrow (1)$ still eludes me at this time, though I thought for a minute that I had figured it out. It’s more difficult because additivity is not preserved by translation, unlike convexity. For another instance of this, we could consider the condition “$\partial K_x$ is additive for all $x\in V$.” This is never true when $\dim V>0$, because if $x\neq 0$, then $x\in\partial K_{2x}$, but $2x$ is in the interior of $K_{2x}$.

Answer 3

We wish to investigate the statement $$ \|y-x\|\le\|y\|\land\|z-x\|\le\|z\|\implies\|y+z-x\|\le\|y+z\|\tag0 $$

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Inner Product Spaces

Suppose that we are in an inner product space: $|u|^2=\langle u,u\rangle$.

Proposition 1: $$ |y-x|\le|y-o|\land|z-x|\le|z-o|\implies\left|\frac{y+z}2-x\,\right|\le\left|\frac{y+z}2-o\,\right|\tag1 $$ Proof: $$ \begin{align} &\left|\frac{y+z}2-o\,\right|^2-\left|\frac{y+z}2-x\,\right|^2\\[6pt] &=\langle(y+z)-(x+o),x-o\rangle\tag{2a}\\[6pt] &=\left\langle y-\frac{o+x}2,x-o\right\rangle+\left\langle z-\frac{x+o}2,x-o\right\rangle\tag{2b}\\ &=\frac12\left(|y-o|^2-|y-x|^2\right)+\frac12\left(|z-o|^2-|z-x|^2\right)\tag{2c}\\[6pt] &\ge0\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: $\langle u+v,u-v\rangle=|u|^2-|v|^2$
$\text{(2b)}$: linearity of the inner product
$\text{(2c)}$: $\langle u+v,u-v\rangle=|u|^2-|v|^2$
$\text{(2d)}$: assumptions

$\large\square$

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Proposition 2: $$ |p-x|\le|p-o|\implies\left|p-\frac{x+o}2\,\right|\le|p-o|\tag3 $$ Proof: $$ \begin{align} \left|p-\frac{x+o}2\,\right|^2 &=\left\langle p-\frac{x+o}2,p-\frac{x+o}2\right\rangle\tag{4a}\\ &=\left\langle\frac{p-x}2+\frac{p-o}2,\frac{p-x}2+\frac{p-o}2\right\rangle\tag{4b}\\ &=\frac14\langle p-x,p-x\rangle+\frac12\langle p-x,p-o\rangle+\frac14\langle p-o,p-o\rangle\tag{4c}\\ &\le\frac14|p-x|^2+\frac12|p-x||p-o|+\frac14|p-o|^2\tag{4d}\\[4pt] &\le|p-o|^2\tag{4e} \end{align} $$ Explanation:
$\text{(4a)}$: write norm in terms of inner product
$\text{(4b)}$: rewrite the terms in the inner product
$\text{(4c)}$: use the linearity of the inner product
$\text{(4d)}$: $\langle u,v\rangle\le|u||v|$
$\text{(4e)}$: $|x-p|\le|o-p|$

$\large\square$

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Statement $\textbf{(0)}$ Holds in an Inner Product Space

In general, a counterexample is hard to visualize because Statement $(0)$ is true in an inner product space: $$ \begin{align} \overbrace{\color{#090}{|y-x|\le|y-o|}}^{|y-x|\le|y|}\land\overbrace{\color{#00F}{|z-x|\le|z-o|}}^{|z-x|\le|z|} &\implies\left|\frac{y+z}2-x\,\right|\le\left|\frac{y+z}2-o\,\right|\tag{5a}\\ &\implies\underbrace{\color{#C00}{\left|\frac{y+z}2-\frac{x+o}2\,\right|\le\left|\frac{y+z}2-o\right|}}_{|y+z-x|\le|y+z|}\tag{5b} \end{align} $$ Explanation:
$\text{(5a)}$: Proposition 1
$\text{(5b)}$: Proposition 2 with $p=\frac{y+z}2$

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Importance of Convexity

Proposition 1 says that the region which is closer to $x$ than to $o$ (the origin) is convex. Since $y$ and $z$ are in that region (to the right of the vertical black line), so is $\frac{y+z}2$.

Proposition 2 says that any point that is closer to $x$ than to $o$ is closer to $\frac{x+o}2$ than to $o$. That is, the region to the right of the vertical black line is also to the right of the vertical violet line, and anything to the right of the vertical violet line is closer to $\frac{x+o}2$ than to $o$.

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Other Norms

For inner product spaces, convexity plays an important part in proving Statement $(0)$. Using the $\|\cdot\|_\infty$ norm, $$ \|(x,y)\|_\infty=\max(|x|,|y|)\tag6 $$ the region where $\|p-x\|\le\|p\|$ need not be convex, and the statement fails.

The violet region is where $\|p-x\|\le\|p\|$ and the green region is where $\|p-x/2\|\gt\|p\|$. Since $y$ and $z$ are in the violet region, we have $$ \|y-x\|\le\|y\|\land\|z-x\|\le\|z\|\tag7 $$ However, since $\frac{y+z}2=(0,0)$ is in the green region, we have $$ \|y+z-x\|\gt\|y+z\|\tag8 $$ Inequalities $(7)$ and $(8)$ show that $(0)$ fails.