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

Question

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&#039;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

Jamarcus Shields · Accepted Answer

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&#039;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 · Accepted Answer

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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