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

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The discussion centers on the mathematical relationship between the supremum of the dot product of vectors a and u, constrained by the 2-norm of u, and the expression r times the 2-norm of a. It is established that sup{a·u | ||u||_2 <= r} equals r ||a||_2, which can be demonstrated using the Cauchy-Schwarz inequality. Participants clarify that the 2-norm refers to the Euclidean norm in finite-dimensional spaces. A request for definitions of various norms, including the 2-norm, is made, highlighting the need for foundational understanding in this context. The conversation emphasizes the importance of these mathematical concepts in proving the stated inequality.
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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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