I'm trying to solve a problem that is similar to the fundamental theorem of calculus of variations.Suppose ϕ is a positive, continuous function on [ a , b ] such that ϕ 1 / n → 1 pointwise. Suppose also that f and g are integrable on [ a , b ] such that for all n, ∫ [ a , b ] ( f − g ) ϕ 1 / n = 0. ( ∗ )Show that f = g almost everywhere.My idea is to show that ∫ [ a , b ] ( f − g ) 2 = 0, from which the result follows. So I have, by Fatou's lemma: ∫ [ a , b ] ( f − g ) 2 = ∫ [ a , b ] ( f − g ) 2 lim n → ∞ ϕ 1 / n ≤ lim inf n → ∞ ∫ [ a , b ] ( f − g ) 2 ϕ 1 / n And then I'm stuck at this point since I'm not sure how to evaluate this liminf. I want to say it is equal to 0, but I can't justify it.Also, I haven't used above the fact (*). Any hints are appreciated. Thanks.

Question

I&#039;m trying to solve a problem that is similar to the fundamental theorem of calculus of variations.Suppose   ϕ is a positive, continuous function on   [  a  ,  b  ] such that       ϕ          1              /            n        →  1 pointwise. Suppose also that   f and   g are integrable on   [  a  ,  b  ] such that for all   n,      ∫          [      a      ,      b      ]        (  f  −  g  )        ϕ          1              /            n        =  0.  (  ∗  )Show that   f  =  g almost everywhere.My idea is to show that       ∫          [      a      ,      b      ]        (  f  −  g      )    2    =  0, from which the result follows. So I have, by Fatou&#039;s lemma:      ∫          [      a      ,      b      ]        (  f  −  g      )    2    =      ∫          [      a      ,      b      ]        (  f  −  g      )    2          lim          n      →      ∞            ϕ          1              /            n        ≤      lim inf          n      →      ∞            ∫          [      a      ,      b      ]        (  f  −  g      )    2        ϕ          1              /            n      And then I&#039;m stuck at this point since I&#039;m not sure how to evaluate this liminf. I want to say it is equal to 0, but I can&#039;t justify it.Also, I haven&#039;t used above the fact (*). Any hints are appreciated. Thanks.

Alexia Hart · Accepted Answer

The statement is false. Consider   [  a  ;  b  ]  =  [  −  1  ;  1  ],   ϕ  ≡  1,   f  ≡  0 and   g  (  x  )  =  x. All the conditions are met, however,   f  ≠  g almost everywhere. Presumably it was meant that the conditions hold for all intervals and not a fixed one.In fact, for   [  a  ;  b  ]  =  [  −  1  ;  1  ], we can pick more generally   ϕ  ≡  c  &amp;gt;  0 and   f  (  x  )  =  g  (  −  x  ) (and   f not even), then we still have a counterexample. And if we wish, we can make the same construction on your favourite compact, nontrivial interval by a suitable change of variables.

I'm trying to solve a problem that is similar to the fundamental theorem of calculus of variations.

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