Dual norm

1 Dual norm

p-norm and q-norm are dual if this holds

Let \Vert x\Vert be the norm in the primal space x \in S \subseteq \mathbb{R}^n, then the following expression defines dual norm:

\Vert y\Vert _\star = \sup\limits_{\Vert x\Vert \leq 1} \langle y,x\rangle

The intuition for the finite-dimensional space is how the linear function (element of the dual space) f_y(\cdot) could stretch the elements of the primal space with respect to their size, i.e. \Vert y\Vert _* = \sup\limits_{x \neq 0} \dfrac{\langle y,x\rangle}{\Vert x\Vert }.

2 Properties

  • One can easily define the dual norm as:

    \Vert y\Vert _* = \sup\limits_{x \neq 0} \dfrac{\langle y,x\rangle}{\Vert x\Vert }

  • The dual norm is also a norm itself

  • For any x \in E, y \in E^*: x^\top y \leq \Vert x\Vert \cdot \Vert y\Vert _*

  • \left(\Vert x\Vert _p\right)_* = \Vert x\Vert _q if \dfrac{1}{p} + \dfrac{1}{q} = 1, where p, q \geq 1


The Euclidian norm is self dual \left(\Vert x\Vert_2\right)_\star = \Vert x\Vert _2.

3 Examples


Let f(x) = \Vert x\Vert. Prove that f^\star(y) = \mathbb{O}_{\Vert y\Vert _\star \leq 1}

  1. By definition of the conjugate function:

    f^*(y) = \sup_{x} \{ \langle y, x \rangle - f(x) \} = \sup_{x} \{ \langle y, x \rangle - \|x\| \}

  2. Consider the case \|y\|_* > 1. By definition of the dual norm,

    \Vert y\Vert _* = \sup\limits_{\Vert x\Vert \leq 1} \langle y,x\rangle > 1

    Which means, that there is some x^\dagger, such that \|x^\dagger\|\leq 1, but \langle y,x^\dagger\rangle > 1. Now consider the vector \bar{x} = tx^\dagger, where t \in \mathbb{R}^+. The value of the conjugate function is a supremum, therefore we have the following relation:

    \begin{split} f^*(y) &\geq \langle y, \bar{x} \rangle - \|\bar{x}\| = \langle y, tx^\dagger \rangle - t\|x^\dagger\|\\ &= t(\langle y, x^\dagger \rangle - \|x^\dagger\|) \to \infty \text{ with } t \to \infty \end{split}

    Thus, \|y\|_* > 1 does not belong to the \text{dom } f^*.

  3. Consider the case \|y\|_* \leq 1. By CBS inequality:

    \langle y, x \rangle \leq \| y \|_* \| x \| \leq \| x \|

    Equality holds when x=0. Therefore

    f^*(y) = \sup_{x} \{ \langle y, x \rangle - \|x\| \} = 0