Let μ be a positive Borel measure on (0,1) and g ∈ L 2 ( μ ) be such that ‖ g ‖ 2 = 1.. Then does there exist f ∈ C 0 ( ( 0 , 1 ) ) such that ∫ ( 0 , 1 ) | g | 2 ( f − 1 ) d μ = 0 ?If we can show that the above integral is zero iff f − 1 = 0 almost everywhere then by continuity of f we have f ≡ 1 on (0,1). But then f ∉ C 0 ( ( 0 , 1 ) ) for otherwise it approaches to 0 near the endpoints which is not the case.

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Let   μ be a positive Borel measure on (0,1) and   g  ∈      L    2    (  μ  ) be such that   ‖  g      ‖    2    =  1.. Then does there exist   f  ∈      C    0    (  (  0  ,  1  )  ) such that       ∫          (      0      ,      1      )            |    g            |        2    (  f  −  1  )     d  μ  =  0    ?If we can show that the above integral is zero iff   f  −  1  =  0 almost everywhere then by continuity of   f we have   f  ≡  1 on (0,1). But then   f  ∉      C    0    (  (  0  ,  1  )  ) for otherwise it approaches to 0 near the endpoints which is not the case.

Aibling6n2re · Accepted Answer

I suppose       C    0    (  0  ,  1  ) stands for continuous functions on (0,1) which tend to 0 as   x  →  0 or   x  →  1. Take   f  (  x  )  =  c  x  (  1  −  x  ) You can quickly solve the given equation for   c. Note that       ∫    0          1        x  (  1  −  x  )      |    g  (  x  )            |              2        d  x  ≠  0.[ We need   c      ∫    0          1        x  (  1  −  x  )      |    g  (  x  )            |              2        d  μ  =  ∫      |    g            |              2        d  μ which is satisfied when   c  =            ∫              |            g                        |                          2                    d      μ                      ∫        0                  1                    x      (      1      −      x      )              |            g      (      x      )                        |                          2                    d      μ      ]

Let μ be a positive Borel measure on (0,1) and g ∈

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