Showing f(x-y) = f(x) - f(y) for Additive Functions

  • Thread starter Ed Quanta
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In summary, we can prove that if f(x+y)=f(x)+f(y) for all x and y in the Reals in p space, and f(0)=0, then f(x-y)=f(x)-f(y). Additionally, if f is continuous at some point c in the Reals in p space, it implies that f is continuous throughout the entire Reals in p space. This can be shown by translating everything back to c and using the definition of continuity. Furthermore, it is possible to prove that if f(x+y)=f(x)+f(y) and f is bounded on any interval, no matter how small, then f is continuous. This can be proven by considering Cauchy sequences and using the
  • #1
Ed Quanta
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Ok, so suppose f is a function which takes us from the Reals in p space to the Reals in m space. And f(x+y)=f(x) + f(y) for all x and y 's in the Reals in p space.

Now if f(0)=0, how do I now show f(x-y)= f(x) - f(y)?
 
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  • #2
f(y-y) = f(0) = f(y) + f(-y) = 0 therefore f(-y) = -f(y).
 
  • #3
Ahhhh, so now if f is continuous at some c is an element of the Reals in p space, how does this imply f is continuous throughout the Reals in p space?
 
  • #4
Because you can translate everything back to c.
 
  • #5
How?

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

so were f(x-y)=f(n), we can always write this in terms of c

f(c-t)=f(c)-f(t) where t=c-n

We know at c, f is continuous, but how do we then conclude that for f(t)?
 
  • #6
Let x= c+t, y=c+s for some s,t consider f(x-t) and the fact that if |x-y| < d then |t-s| < d

or if you prefer, an interval of width d centred on x can be translated back to an interval of width d centred on c.
 
  • #7
Use 0 (the 0 vector in Rn of course).

Knowing that f is continuous at q means
lim(x->q)f(x)= f(q). Let h= x-q so that x= q+h and x= q is equivalent to h= 0. The limit becomes lim(h->0)f(q+h)= lim(h->0) f(q)+ f(h)= f(q)+ lim(h->0)f(h)= f(q) so
lim(h->0)f(h)= 0.

Now, if p is any other point, lim(x->p)f(x)= lim(h->0)f(p+ h) (taking h= x- p)
= lim(h->0) f(p)+ f(h)= f(p)+ lim(h->0)f(h)= f(p)+ 0= f(p) so that f is continuous at p.

In fact, it is possible to prove (but much harder, I understand) that if f(x+y)= f(x)+ f(y) and f is bounded on any interval, no matter how small, then f is continuous.
 
  • #8
Is it hard? I've not seen this one before like this (though it's close to the bounded implies cont at 0)

Let x_n be a cauchy seqence tending to 0. Let y_n be the largest integer such that |x_ny_n| is still less than 1. Passing to a s subsequence, we may assume that y_n is strictly monotone increasing and y_n =>n, but then

|f(x_n)| < M/n where M bounds the value on the unit ball.

so f(x_n) is cauchy, hence converges to zero, so it's continuous at 0 and the previous result applies.
 

1. What is an additive function?

An additive function is a mathematical function that satisfies the property f(x+y) = f(x) + f(y) for all values of x and y. This means that when two values are added together and input into the function, the output will be equivalent to the sum of the individual outputs when each value is input separately.

2. How do you prove that f(x-y) = f(x) - f(y) for additive functions?

To prove this property for an additive function, you can use the definition of an additive function and algebraic manipulations. Start with f(x-y) and substitute in the definition of an additive function, f(x+y) = f(x) + f(y). Then, use algebraic manipulations such as factoring and rearranging terms to show that f(x-y) is equivalent to f(x) - f(y).

3. Can you provide an example of an additive function?

One example of an additive function is the function f(x) = 2x. When we input two values, say 3 and 5, and add them together, we get f(3+5) = f(8) = 16. This is equivalent to the sum of the individual outputs, f(3) + f(5) = 6 + 10 = 16.

4. What is the significance of proving f(x-y) = f(x) - f(y) for additive functions?

This property is important because it allows us to simplify calculations involving additive functions. Instead of having to calculate f(x) and f(y) separately and then add them together, we can simply input the sum of x and y into the function and get the same result. This also helps in proving other properties and theorems involving additive functions.

5. Are all linear functions additive functions?

Yes, all linear functions are additive functions. This is because the definition of a linear function is f(x) = mx + b, and when we substitute in two values for x and add them together, we get f(x+y) = m(x+y) + b = mx + b + my + b = f(x) + f(y). Therefore, linear functions satisfy the property of additive functions.

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