(i) if ` f ( (x + 2y)/3)]= (f(x) +2f(y))/3 , " x, y in R and f ` (0) exists and is finite ,show that f(x) is continuous on the entire number line.
(ii) Let f : ` R to R ` satisfying f(x) -f (y)`| le | x-y|^(3) , AA x, y in R ` , The prove that f(x) is a constant function.

Explanation

(i) we have
`f((x+2y)/3) = (f(x) +2f(y))/3, AA x, y in R and f(x)` is differentiable at x =0
i.e. `f'(0) = lim_(x to 0) (f (x +h)-f(x))/h `
`lim_(h to 0) (f((3x+2.(3h)/2)/3) -f((3x+2.0)/3))/h `
` lim_(h to0) ((f(3x) +2f((3h)/2))/3-(f(3x)+2f(0))/3)/h`
`lim_( h to 0) (2f((3h)/2)-2f(0))/(3h) = lim_(h to 0) (f((3h)/2)-f(0))/((3h)/2) = f'(0) = k ,say `
` Rightarrow f'(x) = k `
` Rightarrow f(x) = k x + c , k, c in R `
Which is a linear function in x, hence f(x) is continuous for all x.
(ii) we have
`| f(x) -f(y) le|x -y|^(3), x ne y`
`Rightarrow |(f(x)-f(y))/(x-y)| le|x-y|^(2)`
` Rightarrow lim_( y tox)|(f(y)-f(x))/(y-x)|le lim_( y to x) |x-y|^(2)`
`Rightarrow |f'(x)|le 0 " but " |f'(x)|cancellt 0`
`Rightarrow |f'(x)|=0`
`Rightarrow f(x)= c ` , ( constant)
` Rightarrow f (x)` is a constant function