Question

Suppose that a function f, defined for all real numbers, satisfies the property that, for all x and y, f(x + y) = f(x) + f(y). Suppose in addition to (*) that f is differentiable at x = 0. Prove that f is differentiable

Answer 1

To prove that f is differentiable, we can use the property f(x+y) = f(x) + f(y) and differentiate both sides. We find that f is differentiable at any real number y, which implies f is differentiable for all real numbers.

Step-by-step explanation:

To prove that f is differentiable, we need to show that the limit lim x→a [f(x)-f(a)]/(x-a) exists as x approaches a for any real number a.

Using the property given in the question that f(x+y) = f(x) + f(y), we can write:

f(x) = f(x+y-y) = f(x+y) - f(y)

Now, let's differentiate both sides with respect to x:

f'(x) = f'(x+y)

Since f is differentiable at x = 0, we have:

f'(0) = f'(y)

Therefore, the function f is differentiable at any real value of y, which implies f is differentiable for all real numbers.