Let ϕ ( s ) := ∫ 0 ∞ exp ⁡ ( − s t ) g ( t ) d t for g ∈ L 1 ( 0 , ∞ ). Assume that ϕ ( s ) = 0 for s ∈ [ 0 , 1 2 ). How to prove that ϕ ( s ) = 0 for every s ∈ [ 0 , ∞ )

Name: Let ϕ ( s ) := ∫ 0 ∞ exp ⁡ ( − s t ) g ( t ) d t for g ∈ L 1 ( 0 , ∞ ). Assume that ϕ ( s ) = 0 for s ∈ [ 0 , 1 2 ). How to prove that ϕ ( s ) = 0 for every s ∈ [ 0 , ∞ )
Uploaded: 2022-02-23T17:22:57+00:00
Description: Let ϕ ( s ) := ∫ 0 ∞ exp ⁡ ( − s t ) g ( t ) d t for g ∈ L 1 ( 0 , ∞ ). Assume that ϕ ( s ) = 0 for s ∈ [ 0 , 1 2 ). How to prove that ϕ ( s ) = 0 for every s ∈ [ 0 , ∞ )

Question

Let   ϕ  (  s  )  :=      ∫          0              ∞        exp  ⁡  (  −  s  t  )  g  (  t  )  d  t for   g  ∈      L    1    (  0  ,  ∞  ). Assume that   ϕ  (  s  )  =  0 for   s  ∈  [  0  ,      1    2    ). How to prove that   ϕ  (  s  )  =  0 for every   s  ∈  [  0  ,  ∞  )

Aydin Rodgers · Accepted Answer

I claim that   ϕ is an entire function.Indeed let   γ be a simply closed curve in the complex plane.Then       ∫          γ        ϕ  (  z  )  d  z  =      ∫          γ            ∫    0    ∞    exp  ⁡  (  −  z  t  )  g  (  t  )  d  z  d  t  =      ∫    0    ∞        ∫    γ    exp  ⁡  (  −  z  t  )  g  (  t  )  d  z  d  t by Fubini&#039;s theorem.But the function   exp  ⁡  (  −  z  t  ) is entire as a function of z.Hence, by the Cauchy-Goursat theorem       ∫    γ    exp  ⁡  (  −  z  t  )  d  z vanishes for all   t  ∈  (  0  ,  ∞  )Therefore,       ∫    γ    ϕ  (  z  )  d  z  =  0 for every simply closed curve   γ in the complex plane. Moreras theorem, then, implies that   ϕ is entire.Now since   ϕ is an analytic function that vanishes on a set with accumulation point we get that   ϕ is identically zero by the uniqueness theorem for analytic functions.

Let phi(s):=int_0^(oo) exp(-st)g(t)dt for g in L_1(0,oo). Assume that phi(s)=0 for s in [0,1/2). How to prove that phi(s)=0 for every s in [0,oo).

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