I am strugling with this exercise:Let f n ( t ) = sin ⁡ ( n t ). Prove that f n ⇀ ∗ 0 in L ∞ [ 0 , 2 π ]The solution given to me reads:Since the given space is the dual of L 1 [ 0 , 2 π ] : L ∞ [ 0 , 2 π ] = ( L 1 [ 0 , 2 π ] ) ′ Thinking of the f n ( t ) as linear functionals with domain in L 1 [ 0 , 2 π ]Prove that f n ( g ) = ∫ 0 2 π g ( t ) f n ( t ) d t → 0 for all g ∈ L 1 [ 0 , 2 π ]It is enough to prove it holds for g in a set whose span is dense in L 1 [ 0 , 2 π ], like the Δ = { 1 [ a , b ] : [ a , b ] ⊂ [ 0 , 2 π ] }.That is straightforward since I can easily integrate sin(nt) over [a,b]and notice it converges to 0, but I don't understand why/ I am not convinced that it is enough to do so. I have read elsewhere that " a sequence of continuous functions on a metric space that converges pointwise on a dense subset need not converge pointwise on the full space.",So why does it work in this particular case? Can you prove it?I have the following proven propositions, but none of them are exactly what I need, since this problem deals with weak * convergence insteadProposition 1: X: Banach space, E ⊂ X, such that ( S p a n E ) ¯ = X. Let { g n } ⊂ X ′ , bounded such that ∃ lim n → ∞ g n ( x ) ∀ x ∈ E. Then ∃ g ∈ X ′ such that lim n → ∞ g n ( x ) = g ( x ) ∀ x ∈ Xwhich was actually used to prove:Proposition 2: X: Banach space, x n ⇀ x iff { x n } is bounded and lim n → ∞ f ( x n ) = f ( x ) ∀ f ∈ Δ ⊂ X ′ 'such that ( S p a n Δ ) ¯ = X ′

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I am strugling with this exercise:Let       f    n    (  t  )  =  sin  ⁡  (  n  t  ). Prove that       f    n    ⇀  ∗  0 in       L          ∞        [  0  ,  2  π  ]The solution given to me reads:Since the given space is the dual of       L    1    [  0  ,  2  π  ]  :       L          ∞        [  0  ,  2  π  ]  =  (      L    1    [  0  ,  2  π  ]      )    ′  Thinking of the       f    n    (  t  ) as linear functionals with domain in       L    1    [  0  ,  2  π  ]Prove that       f    n    (  g  )  =      ∫    0          2      π        g  (  t  )      f    n    (  t  )  d  t  →  0 for all   g  ∈      L    1    [  0  ,  2  π  ]It is enough to prove it holds for   g in a set whose span is dense in       L    1    [  0  ,  2  π  ], like the   Δ  =  {      1          [      a      ,      b      ]        :  [  a  ,  b  ]  ⊂  [  0  ,  2  π  ]  }.That is straightforward since I can easily integrate sin(nt) over [a,b]and notice it converges to 0, but I don&#039;t understand why/ I am not convinced that it is enough to do so. I have read elsewhere that &quot; a sequence of continuous functions on a metric space that converges pointwise on a dense subset need not converge pointwise on the full space.&quot;,So why does it work in this particular case? Can you prove it?I have the following proven propositions, but none of them are exactly what I need, since this problem deals with weak * convergence insteadProposition 1: X: Banach space,   E  ⊂  X, such that             (      S      p      a      n      E      )        ¯    =  X. Let   {      g    n    }  ⊂      X    ′  , bounded such that   ∃      lim          n      →      ∞            g    n    (  x  )   ∀  x  ∈  E. Then   ∃  g  ∈      X    ′   such that       lim          n      →      ∞            g    n    (  x  )  =  g  (  x  )   ∀  x  ∈  Xwhich was actually used to prove:Proposition 2: X: Banach space,       x    n    ⇀  x iff   {      x    n    } is bounded and       lim          n      →      ∞        f  (      x    n    )  =  f  (  x  )   ∀  f  ∈  Δ  ⊂      X    ′   &#039;such that             (      S      p      a      n      Δ      )        ¯    =      X    ′

basquinas6v · Accepted Answer

The convergence  ∫  g  (  t  )  sin  ⁡  (  n  t  )  d  t  →  0for   g integrable is known as the Riemann-Lebesgue lemma. There are many proofs of this lemma. I suggest that you look at these proofs, some indeed proceed by density.

I am strugling with this exercise: Let f n </msub> ( t ) = sin

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