I was wondering if the spectral norm is a Lipschitz function with respect to the spectral norm. How can we prove whether it is or not? In other words, is $$\big| \|X\| - \|Y\| \big| \le L \|X-Y\|$$ for some $L$?

  • 1
    $\begingroup$ I'm not familiar with spectral norm, but if it is a norm, can't you just use the triangle inequality to get $|\|X\| - \|Y\|| \leq \|X - Y\|$? $\endgroup$ Sep 28, 2013 at 0:13
  • $\begingroup$ Yes. Thanks. We can use the triangle inequality. $\endgroup$
    – John
    Oct 14, 2013 at 13:41

1 Answer 1


Any norm is Lipschitz of rank 1 with respect to itself. This follows from the traingle inequality.

Since $\|X\| \le \|Y\|+\|X-Y\|$ we have $\|X\| -\|Y\| \le \|X-Y\|$. Reversing the roles of $X,Y$ gives $\|Y\| -\|X\| \le \|X-Y\|$ and combining these gives $| \|X\|-\|Y\| | \le \|X-Y\|$.


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