# Given function f with property f(n)>= (f(n+1)+f(n-1))/2, n integer, show that f is constant if domain is integer Z and range is natural N.

Asked on by luvgoj

sciencesolve | Teacher | (Level 3) Educator Emeritus

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The problem provides the following inequality as being the property of the function, such that:

`f(n) >= (f(n + 1) + f(n - 1))/2 => 2f(n) >= f(n + 1) + f(n - 1)`

You may replace `2f(n)` by `f(n) + f(n)` , such that:

`f(n) + f(n) >= f(n + 1) + f(n - 1) => f(n) - f(n - 1) >= f(n+1) - f(n)`

You may consider the following new function `g(n) = f(n+1) - f(n)` , such that:

`g(n - 1) >= g(n) => g(n - k) >= g(n) >= g(n + k), bar(k,n) in N`

By the assumption made that `g(n) = f(n+1) - f(n)` , yields:

`f(n - k + 1) - f(n - k) >= g(n) >= f(n + k + 1) - f(n + k)`

Let k = 1, such that:

`f(n - 1 + 1) - f(n - 1) >= g(n) >= f(n + 1 + 1) - f(n + 1)`

`f(n) - f(n - 1) >= g(n) >= f(n + 2) - f(n + 1)`

`f(n) >= g(n) >= f(n + 2)`

Let k = 2, such that:

`f(n - 2 + 1) - f(n - 2) >= g(n) >= f(n + 2 + 1) - f(n + 2)`

`f(n - 1) - f(n - 2) >= g(n) >= f(n + 3) - f(n + 2)`

`f(n - 1) >= g(n) >= f(n + 3)`

.......

Let `k = m:`

`f(n - m + 1) - f(n - m) >= g(n) >= f(n + m + 1) - f(n + m)`

Evaluating the following summation, yields:

`f(n) + f(n - 1) + .... + f(n - m + 1) >= m*g(n) >= f(n + 2) + f(n + 3) +... f(n + m + 1) - f(n + m)`

The problem provides the information that the range of the function `f(n)` is the natural set N, `hence f(n) >= 0` , such that:

`-f(n+1) <= m*g(n) <= f(n) => -(1/m)f(n+1) <= g(n) <= (1/m)f(n)`

`m in N`

If m increases, then `1/m -> 0 => g(n) = 0 => f(n+1) - f(n) = g(n) = 0 => f(n) = f(n+1)` .

Hence, evaluating if the given function f(n) is constant, under the given conditions, yields `f(n) = f(n+1)` that proves that f(n) is constant for all `n in Z` .

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