Let ϕ 1 , . . . , ϕ k be linear functionals on R...

Addison Trujillo

Addison Trujillo

Answered

2022-07-13

Let ϕ 1 , . . . , ϕ k be linear functionals on R n , c 1 , . . . c k R and
H = { x R n | | ϕ j ( x ) | c j , j = 1 , . . , n }
Under what conditions is H compact? So my intuition tells me we must have k n, and ϕ 1 , . . . , ϕ k and at least n of them must be linearly independent in the dual space, not sure how to prove it though, since we're working with a system of inequalities rather then equalities. That is, my idea was to write:
| ϕ 1 ( x ) | c 1 | α 11 x 1 + . . . + α 1 n x n | c 1
| ϕ 2 ( x ) | c 2 | α 21 x 1 + . . . + α 2 n x n | c 2
...
| ϕ k ( x ) | c k | α k 1 x 1 + . . . + α k n x n | c k

Answer & Explanation

Jamarcus Shields

Jamarcus Shields

Expert

2022-07-14Added 17 answers

Hint: The set H is a closed subset of R n by construction, so by the Heine Borel theorem, it is compact if and only if it is bounded. This is the case if and only if the functionals ϕ 1 , , ϕ k span the dual space. If they don't span the dual, then they have a non-zero common kernel and that common kernel is contained in H which thus is unbounded. On the other hand, if they span the dual, then one easily concludes that the image of H under each linear functional is bounded, which easily implies that H is bounded.
ScommaMaruj

ScommaMaruj

Expert

2022-07-15Added 5 answers

Take the elements of the dual of the standard basis, so ϕ i ( x ) = x i for x = ( x 1 , , x n ). So you get information like that x H implies | x i | a i for some constant ai and all i = 1 , , n and this implies that H is bounded.

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