Let f : R k → R be a bounded harmonic function (i.e. f is of class C 2 and ∑ r = 1 k ∂ 2 f ∂ y r 2 = 0) and ( x 1 , . . . , x k ) ∈ R k .Prove that f ( x 1 , . . . , x k ) = 1 ( 2 π ) k ∫ R k f ( x 1 + y 1 , . . . , x k + y k ) e − 1 2 ∑ r = 1 k y r 2 d y 1 . . . d y k .We note that ∫ R k f ( x 1 + y 1 , . . . , x k + y k ) e − 1 2 ∑ r = 1 k y r 2 d y 1 . . . d y k = ∫ R k f ( y 1 , . . . , y k ) e − 1 2 ∑ r = 1 k ( y r − x r ) 2 d y 1 . . d y k ,how to use the fact that f is harmonic ?

Question

Let   f  :            R        k    →      R   be a bounded harmonic function (i.e.   f is of class       C    2   and       ∑          r      =      1        k                      ∂        2            f              ∂              y        r        2              =  0) and   (      x    1    ,  .  .  .  ,      x    k    )  ∈            R        k    .Prove that  f  (      x    1    ,  .  .  .  ,      x    k    )  =      1          (              2        π                    )        k                  ∫                            R                k              f  (      x    1    +      y    1    ,  .  .  .  ,      x    k    +      y    k    )      e          −              1        2                    ∑                  r          =          1                k                    y        r        2              d      y    1    .  .  .  d      y    k    .We note that      ∫                            R                k              f  (      x    1    +      y    1    ,  .  .  .  ,      x    k    +      y    k    )      e          −              1        2                    ∑                  r          =          1                k                    y        r        2              d      y    1    .  .  .  d      y    k    =      ∫                            R                k              f  (      y    1    ,  .  .  .  ,      y    k    )      e          −              1        2                    ∑                  r          =          1                k            (              y        r            −              x        r                    )        2              d      y    1    .  .  d      y    k    ,how to use the fact that   f is harmonic ?

Kaylie Mcdonald · Accepted Answer

Note: I&#039;ve used   n instead of   k - sorry this is a force of habit. Since   u is harmonic it satisfies the mean value property:  u  (  x  )  =      1          n              ω        n                    r                  n          −          1                          ∫          ∂              B        r            (      x      )        u  (  y  )    d  S    for all   r  &amp;gt;  0where       ω    n   is the volume of the   n-ball with radius 1. Hence,  n      ω    n        r          n      −      1        u  (  x  )  =      ∫          ∂              B        r            (      x      )        u  (  y  )    d  S    for all   r  &amp;gt;  0.Multiply through by       e          −              1        2                    r        2             on both side then integrate from 0 to   ∞. On the left hand side you get  n      ω    n        ∫    0    ∞    u  (  x  )      r          n      −      1            e          −              1        2                    r        2                d  r  =  n      ω    n    u  (  x  )      ∫    0    ∞        r          n      −      1            e          −              1        2                    r        2                d  r  =  n      ω    n    u  (  x  )            Γ      (      n              /            2      )              2              1        +        n                  /                2              .It is known that       ω    n    =            π              n                  /                2                    Γ      (      n              /            2      +      1      )      . Hence the LHS is:  n  u  (  x  )            π              n                  /                2                    Γ      (      n              /            2      +      1      )        ⋅            Γ      (      n              /            2      )              2              1        +        n                  /                2              =  u  (  x  )  (  2  π      )          n              /            2      using properties of the gamma function. On the right hand side you get      ∫    0    ∞        ∫          ∂              B        r            (      x      )        u  (  y  )      e          −              1        2                    r        2                d  S  d  r  =      ∫                            R                n              u  (  y  )      e          −              1        2            |      y      −      x              |        2                d  yvia polar coordinates. Thus,  u  (  x  )  =      1          (      2      π              )                  n                      /                    2                          ∫                            R                n              u  (  y  )      e          −              1        2            |      y      −      x              |        2                d  y  .

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