# Dunkl-Williams inequality

Let $V$ be an inner product space  and $a,b\in V$. If $a\not=0$ and $b\not=0$, then

 $\|a-b\|\geq\frac{1}{2}(\|a\|+\|b\|)\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\|.$ (1)

Equality holds if and only if $a=0$, $b=0$, $\|a\|=\|b\|$ or $a\|b\|=b\|a\|$. In fact, if (1) holds and $V$ is a normed linear space, then $V$ is an inner product space.

If $X$ is a normed linear space and $a\not=0$ and $b\not=0$ then

 $\|a-b\|\geq\frac{1}{4}(\|a\|+\|b\|)\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\|.$ (2)

Equality holds if and only if $a=0$, $b=0$ or $a=b$. The constant $\frac{1}{4}$ is best possible. For example, let $X$ be the set of ordered pairs of real numbers, with norm of $(x_{1},x_{2})$ equal to $|x_{1}|+|x_{2}|.$ Let $a=(1,\epsilon)$ and $b=(1,0)$ where $\epsilon$ is a small positive number. After a bit of routine calculation, it is easily seen that the best possible constant is $\frac{1}{4}$.

The inequality  (2) has been generalized in the case where $X$ is a normed linear space over the reals. In that case one can show:

 $\|a-b\|\geq c_{p}(\|a\|^{p}+\|b\|^{p})^{1/p}\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\|$ (3)

where $c_{p}=2^{-1-1/p}$ if $0 and $c_{p}=1/4$ if $p\geq 1$. The case $p=1$ is the Dunkl and Williams inequality.

If $X$ is a normed linear space and $0 then (3) holds with $c_{p}=2^{-1/p}$ if and only if $X$ is an inner product space.

The inequality (2) can be improved slightly to get:

 $\|a-b\|\geq\frac{1}{2}\max(\|a\|,\|b\|)\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\|.$ (4)

Equality holds in (4) if and only if $a$ and $b$ span an ${\ell_{2}}^{1}$ in the underlying real vector space with $\pm\|b-a\|^{-1}(b-a)$ and $\pm\|a\|^{-1}a$ (or $\pm\|b\|^{-1}b$) as the vertices of the unit parallelogram.

## References

Title Dunkl-Williams inequality DunklWilliamsInequality 2013-03-22 16:56:38 2013-03-22 16:56:38 Mathprof (13753) Mathprof (13753) 12 Mathprof (13753) Definition msc 47A12