# Norm inequality (supper bound)

Do you think this inequality is correct? I try to prove it, but I cannot. Please hep me. Assume that $\|X\| < \|Y\|$, where $\|X\|, \|Y\|\in (0,1)$ and $\|Z\| \gg \|X\|,\|Z\| \gg \|Y||$. prove that $$\|X+Z\|-\|Y+Z\| \leq \|X\|-\|Y\|$$ and if $Z$ is increased, the left hand side become smaller. I pick up some example and see that this inequality is correct but I cannot prove it. Thank you very much.

-

## migrated from mathoverflow.netJul 7 '13 at 18:41

This question came from our site for professional mathematicians.

I only need to prove with X,Y,Z∈R^2, I'm sorry for this inconvenience, because I'm an engineer. –  Trinh Chien Jul 7 '13 at 7:01

The inequality is false as stated. Let

\begin{align} X &= (0.5,0)\\ Y &= (-0.7,0)\\ Z &= (z,0), 1 \ll z \end{align}

This satisfies all the conditions given. We have that

$$\|X + Z\| - \|Y + Z\| = z + 0.5 - (z - 0.7) = 1.2 \not\leq -0.2 = \|X\| - \|Y\|$$

From the Calculus point of view, in $n$ dimensions, we can write

$$\|X\| = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}$$

We have that when $\|X\| \ll \|Z\|$, we can approximate $\|X+Z\| \approx \|Z\| + X\cdot \nabla(\|Z\|)$. Now, $\nabla(\|Z\|) = \frac{Z}{\|Z\|}$ by a direct computation, so we have that

$$\|X + Z\| - \|Y + Z\| \approx (X-Y) \cdot \frac{Z}{\|Z\|}$$

From this formulation we see that even in the cases where $X,Y$ are infinitesimal the inequality you hoped for cannot hold true. However, the right hand side of this approximation can be controlled by Cauchy inequality to get (using that $Z / \|Z\|$ is a unit vector).

$$(X-Y) \cdot \frac{Z}{\|Z\|} \leq \|X - Y\|$$

So perhaps what you are thinking about is the following corollary of the triangle inequality

Claim: If $X,Y,Z$ are vectors in $\mathbb{R}^n$, then $$\|X + Z\| - \|Y + Z\| \leq \|X - Y \|$$

Proof: We write $$X + Z = (X - Y) + (Y + Z)$$ so by the triangle inequality $$\|X + Z\| = \|(X - Y) + (Y+Z)\| \leq \|X - Y\| + \|Y + Z\|$$ rearranging we get $$\|X + Z\| - \|Y + Z\| \leq \|X - Y\|$$ as desired.

Remark: if we re-write the expression using $-Z$ instead of $Z$, the same claim is true in an arbitrary metric space: Let $(S,d)$ be a metric space. Let $x,y,z$ be elements of $S$. Then $$d(x,z) - d(y,z) \leq d(x,y)$$.

-
Sorry for this inconvenience, but Could I ask you a question,prof. Willie Wong? is this equation correct? ∥X+Z∥−∥Y+Z∥≈(X−Y).(Z/‖Z‖) As I see, ∥X+Z∥−∥Y+Z∥ is a number and (X−Y).(Z/∥Z∥) is a matrix? I'm sorry if it is a stupid question. –  Hoping_VN Jul 8 '13 at 11:11
$Z$ is a vector. $(X-Y)$ is a vector. Their dot product $(X-Y)\cdot Z$ is a scalar. –  Willie Wong Jul 8 '13 at 11:31