function f:R->R can be written as a sum f=f1+f2 where f1 is even and f2 is odd。then if f is continuous then f1 and f2 may be chosen continuous, and if f is differentiable then f1 and f2 can be chosen differentiable i am quiet confusing this statement , if f1 is continuous f2 is not how their sum to be continuous and differentiable as well. but i am sure this statement is true. can someone explain to me ???
I don't know if you could have [itex]f_1[/itex] continuous everywhere and [itex]f_2[/itex] discontinuous. A simple way that you could have two discontinuous functions sum to a continuous one is if you take cosine from [itex]-\infty<\theta\leq 0[/itex] and 0 when greater than zero as [itex]f_1[/itex] and then a similar thing but where it is defined on the positive number and 0 for the negatives (and at 0 or else it will still be discontinuous). Individually there is a discontinuity at 0, but when you add them they are continuous. This is a trivial case, but I hope it makes it clear.
You are correct that if f1 is continous at, say, x= a, and f2 is not, then f cannot be continuous at x= a. But why are you raising the question? This statement says nothing about "f1 is continuous f2 is not". It simply says that if f is continuous then f1 and f2 can both be chosen to be continuous. Here it is fairly standard to take f1= (f(x)+ f(-x))/2 and f2= (f(x)- f(-x))/2. If f is continuous, so are f1 and f2. If f is differentiable, so are f1 and f2.