Why is Supremum of a.u Less Than or Equal to r||a||_2?

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SUMMARY

The supremum of the dot product of vectors a and u, constrained by the 2-norm of u, is established as less than or equal to r times the 2-norm of a, formally expressed as sup{a.u | ||u||_2 <= r} = r ||a||_2. This relationship is derived from the Cauchy-Schwarz inequality, which provides a foundational proof for this inequality in finite-dimensional spaces. The 2-norm, also known as the Euclidean norm, is crucial for understanding this concept.

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Consider a and u are vector of n entries,
why the supremum of a dot u subject to the 2-norm of u is less than or equal to r equals r times 2-norm of a, i.e. sup{a.u | ||u||_2 <=r} = r ||a||_2?

How can I work out that?

Thank you!
 
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peterlam said:
Consider a and u are vector of n entries,
why the supremum of a dot u subject to the 2-norm of u is less than or equal to r equals r times 2-norm of a, i.e. sup{a.u | ||u||_2 <=r} = r ||a||_2?

How can I work out that?

Thank you!

Can you please give a definition of the two-norm as I haven't encountered it before (1 and infinity norm, and L^p norms but not "2-norm").
 
Sorry. I mean Euclidean norm.

Thanks!
 
chiro said:
Can you please give a definition of the two-norm as I haven't encountered it before (1 and infinity norm, and L^p norms but not "2-norm").
This would be the \ell^2 norm, except that here we are simply in a finite-dimensional space. In more conventional notation,

\sup\{\left&lt;x,y\right&gt;\ :\ \|y\|\leq r\}=r\|x\|.

@peterlam: to show LHS \leq RHS use Cauchy-Schwartz .
 

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