Question

If 'f' be a real value function satisfying f(x+ 3/2) + f(x) = f(x+1) + f(x+1/2) and |f(x)| ≤ 2.
Then prove that f(x) is periodic.


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Answer 1

Answer:

Step-by-step explanation:

take   g[x] =  f[x+1] -  f(x)  

f(x+ 3/2) + f(x) = f(x+1) + f(x+1/2)  

f(x+ 3/2) - f(x + 1/2) = f(x+1) - f(x)

g[x+1/2] = g[x]    

clearly   g[x] is a periodic function of period   1/2    

let       g[0] = k    then g[1/2] =k  and  g[1] =k

g[0] = f[1] -f[0] =k

         f[1] = f[0] +k

g[1] = f[2] -f[1] =k

         f[2] = f[1] +k =f[0] +k +k = f(0) +2k

processing like this

          f[n] = f[0] +nk     for any n

for all x   If(x)I ≤ 2

              nk ≤ 2  for any n   k =0    

       f[n] = f[0]     for any n

f[x] is periodic

           

Answer 2

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Hence proved ......