Is the spectral norm a Lipschitz function with respect to the spectral norm?

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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 $L$?




real-analysis norm lipschitz-functions matrix-norms spectral-norm




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    I'm not familiar with spectral norm, but if it is a norm, can't you just use the triangle inequality to get $||X| - |Y|| leq |X - Y|$?
    – Pratyush Sarkar
    Sep 28 '13 at 0:13










  • Yes. Thanks. We can use the triangle inequality.
    – John
    Oct 14 '13 at 13:41














up vote
0
down vote

favorite












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 $L$?




real-analysis norm lipschitz-functions matrix-norms spectral-norm




share|cite|improve this question


















  • 1




    I'm not familiar with spectral norm, but if it is a norm, can't you just use the triangle inequality to get $||X| - |Y|| leq |X - Y|$?
    – Pratyush Sarkar
    Sep 28 '13 at 0:13










  • Yes. Thanks. We can use the triangle inequality.
    – John
    Oct 14 '13 at 13:41












up vote
0
down vote

favorite









up vote
0
down vote

favorite











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 $L$?




real-analysis norm lipschitz-functions matrix-norms spectral-norm




share|cite|improve this question














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 $L$?





real-analysis norm lipschitz-functions matrix-norms spectral-norm





share|cite|improve this question













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edited Aug 21 at 10:38









Rodrigo de Azevedo

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asked Sep 27 '13 at 23:53









John

7817




7817







  • 1




    I'm not familiar with spectral norm, but if it is a norm, can't you just use the triangle inequality to get $||X| - |Y|| leq |X - Y|$?
    – Pratyush Sarkar
    Sep 28 '13 at 0:13










  • Yes. Thanks. We can use the triangle inequality.
    – John
    Oct 14 '13 at 13:41












  • 1




    I'm not familiar with spectral norm, but if it is a norm, can't you just use the triangle inequality to get $||X| - |Y|| leq |X - Y|$?
    – Pratyush Sarkar
    Sep 28 '13 at 0:13










  • Yes. Thanks. We can use the triangle inequality.
    – John
    Oct 14 '13 at 13:41







1




1




I'm not familiar with spectral norm, but if it is a norm, can't you just use the triangle inequality to get $||X| - |Y|| leq |X - Y|$?
– Pratyush Sarkar
Sep 28 '13 at 0:13




I'm not familiar with spectral norm, but if it is a norm, can't you just use the triangle inequality to get $||X| - |Y|| leq |X - Y|$?
– Pratyush Sarkar
Sep 28 '13 at 0:13












Yes. Thanks. We can use the triangle inequality.
– John
Oct 14 '13 at 13:41




Yes. Thanks. We can use the triangle inequality.
– John
Oct 14 '13 at 13:41










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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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    accepted










    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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      up vote
      0
      down vote



      accepted










      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 $| |X|-|Y| | le |X-Y|$.






      share|cite|improve this answer






















        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        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 $| |X|-|Y| | le |X-Y|$.






        share|cite|improve this answer












        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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        share|cite|improve this answer



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        answered Oct 28 '13 at 5:56









        copper.hat

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