# Thread: check this proof(is there any fallacy)

1. ## check this proof(is there any fallacy)

( f(x) is a polynomial function and when f(x+y)=f(x)+f(y) given-------------(1) f(
solution
f(0)=0 (proved)
now put y=h where h approaches to zero in (1)
we get f(x+h)=f(x)+f(h)
implies f(x+h)-f(x)=f(h)
implies[ f(x+h)-f(x)]/h=f(h)/h
implies f'(x)=f'(h) { as f(0)=0 }

implies f'(x)=k (as f'(0) is constant) here k is a constant
implies f(x)=kx hence proved
is there any fallacy in this proof.

2. I don't understand what is given.

By the way, there is exist a real function not of the form k*x, which satisfies for all real a,b:

f(a+b)=f(a)+f(b)

3. What is given is that f is a polynomial and that, for all x and y, f(x+ y)= f(x)+ f(y).

What is needed for the proof given here is simply that f is differentiable so "f is a polynomial" is sufficient but not necessary.

Yes, ayushdadhwal, that is a valid proof. You have shown that the derivative is a constant and so f(x)= cx.

It is, by the way, possible to prove this without assuming differentiabllity, only continuity. However, there exist non-continuous functions satisfying f(x+ y)= f(x)+ f(y). They, of course, are nothing like "f(x)= cx".

4. Originally Posted by HallsofIvy
What is given is that f is a polynomial and that, for all x and y, f(x+ y)= f(x)+ f(y).

What is needed for the proof given here is simply that f is differentiable so "f is a polynomial" is sufficient but not necessary.

Yes, ayushdadhwal, that is a valid proof. You have shown that the derivative is a constant and so f(x)= cx.

It is, by the way, possible to prove this without assuming differentiabllity, only continuity. However, there exist non-continuous functions satisfying f(x+ y)= f(x)+ f(y). They, of course, are nothing like "f(x)= cx".
can you give me some other examples

5. What do you mean by "example"? An example of what?

If you are referring to my statement that you really only need continuity, what I did was look at the specific kinds of numbers.

The simplest numbers are the "counting numbers" or positive integers. The defining property of the counting numbers is- counting! Or, in other terms, induction. So I prove by induction that if f(x+ y)= f(x)+ f(y), then f(nx)= nf(x) for any x and n any positive integer.

f(1x)= f(x)= 1f(x). Assume that, for some k, f(kx)= kf(x). Then f((k+1)x)= f(kx+ x)= f(kx)+ f(x)= kf(x)+ f(x)= (k+1)f(x).

The next set of numbers are the "whole numbers" which are just the positive integers together with the number 0. And 0 is, of course, the additive identity: x+ 0= x. f(x)= f(x+ 0)= f(x)+ f(0) so that, subtracting f(x) from each side, f(0)= 0. (Which also says, of course, f(0x)= 0f(x).

The set of all integers includes negatives. If n is any positive integer then f((n+ (-n)x)= f(nx)+ f(-nx). but f(n+ (-n))= f(0)= 0 so we have f(nx)+ f(-nx)= 0. From that, f(-nx)= -f(nx)= -(nf(x))= (-n)f(x). So we have shown that, for all integers n and x any real number, f(nx)= nf(x).

The rational numbers are ratios of integers- the difference is that denominator. If n is any non-zero integer, then n(1/n)= 0 so that f(x)= f(n(1/n)x) and that, from just above, is equal to nf((1/n)x). 'That is, f(x)= nf((1/n)x so that f((1/n)x)= (1/n)f(x).

Of course, then, f((m/n)x)= f(m((1/n)x))= mf((1/n)x)= m((1/n)f(x))= (m/n)f(x).

So far, we have shown that if r is any rational number and x is any real number, f(rx)= rf(x) and, taking x= 1, f(r)= rf(1). Writing c= f(1), that gives f(r)= cr.

The reason we need continuity is that the the set of all real numbers cannot be defined "algebraically"- they must be defined "analytically" using some form of a limit process. The simplest way is this- if x is any real number then there exist a sequence of rational numbers converging to x (This is exactly what we mean by saying that we can write a real number in decimal form. If, say, x= 3.141592... then the sequence 3, 3.1, 3.14, 3.131, 3.1415, 3.14159, 3.141592, ... is a sequence of rational numbers (because they are terminating decimals) that converges to x.)

So for any real number, x, there exist $\displaystyle {r_n}$ converging to x: $\displaystyle f(x)= f(\lim r_n)$ and, with f continuous, that is $\displaystyle \lim f(r_n)= \lim cr= c\lim r= cx$.

And, unlike "differentiable", "continuous is a necessary condition. If we remove that condition, there exist functions satisfying f(x+y)= f(x)+ f(y) that do not satisfy f(x)= cx. ]

To do that, observe that f(x+y)= f(x)+ f(y) is one of the two conditions for a "linear transformation" on a vector space. The other is that f(cx)= cf(x) for c any number. Even without continuity, we were able to prove that f(cx)= cf(x) for c any rational number. So think of the real numbers as a vector space over the field of ratioal numbers. We can define a linear transformation by defining it on a basis for the vector space. The real numbers, as a vector space over the rational numbers has infinite dimension (in fact, uncountable) but we can do the following- since the rational numbers are the underlying field, we can take the number "1" as one basis "vector" and, say $\displaystyle \sqrt{2}$ for another, then whatever (independent) numbers you wish after that. Define f(1)= 1, $\displaystyle f(\sqrt{2})= 2$. Every rational number, r, can be written as "r(1)" so that f(r)= rf(1)= r. But $\displaystyle f(\sqrt{2})= 2\ne \sqrt{2}f(1)=\sqrt{2}$.

While all continuous functions that satisfy f(x+y)= f(x)+ f(y) are of the form cx, non-continuous functions are nasty! In fact, it can be shown the the graph of such a function, far from being a straight line, is dense in the plane. That is, given any point in the plane, there exist a point on the graph arbitrarily close to the given point. Think how awful that would be to graph!