MathJax

Thursday, September 24, 2015

Notes on dual norm

Dual norm is defined as
\[ \|z\|_* = \sup \{z^Tx \;|\; \|x\| \leq 1\}\] The dual norm can be interpreted as the operator norm of \(z^T\), interpreted as a \(1\times n\) matrix with the norm \(\|\cdot\|\) on \(\mathbf{R}^n\).

From the definition of dual norm we obtain the inequality
\[ z^Tx \leq \|x\| \|z\|_*\]
The conjugate of any norm \(f(x)=\|x\|\) is the indicator function of the dual norm unit ball
\[f^*(y) = \left\{ \begin{array}
                             00      & \|y\|_* \leq 1 \\
                             \infty     & \text{otherwise}
                         \end{array}
 \right. \]
Proof. If \(\|y\|_* \gt 1\), then by definition of the dual norm, there is a \(z\in \mathbf{R}^n\) with \(\|z\| \leq 1\) and \(y^T z \gt 1\).  Taking \(x=tz\) and letting \(t \rightarrow \infty\), we have
\[ y^T x - \|x\| = t(y^Tz - \|z\|) \rightarrow \infty\] which shows that \(f^*(y) = \infty\).  Conversely, if \(\|y\|_*\leq 1\), then we have \(y^Tx \leq \|x\| \|y\|^*\) for all \(x\), which implies for all \(x\), \(y^Tx - \|x\| \leq 0\).  Therefore \(x=0\) is the value that maximizes \(y^Tx - \|x\|\), with maximum value 0.

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