Using Hahn-Banach theorem, can we extend an operator to have the same operator norm?

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Above is a theorem in p.74 of "Functional Analysis" by Peter Lax.



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Above is theorem 8 of chapter 3.



My question :
Let $X$ be a normed linear space over the real or complex numbers and $Y$ be a subspace, $l$ be a bounded linear functional on $Y$.
If $l:YtomathbbC$ has operator norm $C$, then can it be extended to $X$ so that the extension has the same operator norm as $l$?




real-analysis linear-algebra functional-analysis operator-theory




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  • 2




    Yes, it's possible. Prove it for norm 1 first, then you can easily prove it for arbitrary norms.
    – Math_QED
    Jul 24 at 15:06






  • 2




    The trick is to consider $p(x) := C |x|$ in the sublinear functional form of HBT (theorem 8 of Chapter 3).
    – Rhys Steele
    Jul 24 at 15:25










  • @Math_QED, Rhys Steele Thank you.
    – GouldBach
    Jul 24 at 15:29














up vote
1
down vote

favorite












enter image description here



Above is a theorem in p.74 of "Functional Analysis" by Peter Lax.



enter image description here



Above is theorem 8 of chapter 3.



My question :
Let $X$ be a normed linear space over the real or complex numbers and $Y$ be a subspace, $l$ be a bounded linear functional on $Y$.
If $l:YtomathbbC$ has operator norm $C$, then can it be extended to $X$ so that the extension has the same operator norm as $l$?




real-analysis linear-algebra functional-analysis operator-theory




share|cite|improve this question















  • 2




    Yes, it's possible. Prove it for norm 1 first, then you can easily prove it for arbitrary norms.
    – Math_QED
    Jul 24 at 15:06






  • 2




    The trick is to consider $p(x) := C |x|$ in the sublinear functional form of HBT (theorem 8 of Chapter 3).
    – Rhys Steele
    Jul 24 at 15:25










  • @Math_QED, Rhys Steele Thank you.
    – GouldBach
    Jul 24 at 15:29












up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here



Above is a theorem in p.74 of "Functional Analysis" by Peter Lax.



enter image description here



Above is theorem 8 of chapter 3.



My question :
Let $X$ be a normed linear space over the real or complex numbers and $Y$ be a subspace, $l$ be a bounded linear functional on $Y$.
If $l:YtomathbbC$ has operator norm $C$, then can it be extended to $X$ so that the extension has the same operator norm as $l$?




real-analysis linear-algebra functional-analysis operator-theory




share|cite|improve this question











enter image description here



Above is a theorem in p.74 of "Functional Analysis" by Peter Lax.



enter image description here



Above is theorem 8 of chapter 3.



My question :
Let $X$ be a normed linear space over the real or complex numbers and $Y$ be a subspace, $l$ be a bounded linear functional on $Y$.
If $l:YtomathbbC$ has operator norm $C$, then can it be extended to $X$ so that the extension has the same operator norm as $l$?





real-analysis linear-algebra functional-analysis operator-theory





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 24 at 15:01









GouldBach

3368




3368







  • 2




    Yes, it's possible. Prove it for norm 1 first, then you can easily prove it for arbitrary norms.
    – Math_QED
    Jul 24 at 15:06






  • 2




    The trick is to consider $p(x) := C |x|$ in the sublinear functional form of HBT (theorem 8 of Chapter 3).
    – Rhys Steele
    Jul 24 at 15:25










  • @Math_QED, Rhys Steele Thank you.
    – GouldBach
    Jul 24 at 15:29












  • 2




    Yes, it's possible. Prove it for norm 1 first, then you can easily prove it for arbitrary norms.
    – Math_QED
    Jul 24 at 15:06






  • 2




    The trick is to consider $p(x) := C |x|$ in the sublinear functional form of HBT (theorem 8 of Chapter 3).
    – Rhys Steele
    Jul 24 at 15:25










  • @Math_QED, Rhys Steele Thank you.
    – GouldBach
    Jul 24 at 15:29







2




2




Yes, it's possible. Prove it for norm 1 first, then you can easily prove it for arbitrary norms.
– Math_QED
Jul 24 at 15:06




Yes, it's possible. Prove it for norm 1 first, then you can easily prove it for arbitrary norms.
– Math_QED
Jul 24 at 15:06




2




2




The trick is to consider $p(x) := C |x|$ in the sublinear functional form of HBT (theorem 8 of Chapter 3).
– Rhys Steele
Jul 24 at 15:25




The trick is to consider $p(x) := C |x|$ in the sublinear functional form of HBT (theorem 8 of Chapter 3).
– Rhys Steele
Jul 24 at 15:25












@Math_QED, Rhys Steele Thank you.
– GouldBach
Jul 24 at 15:29




@Math_QED, Rhys Steele Thank you.
– GouldBach
Jul 24 at 15:29















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