# Math Help - Grade my continuity proof

1. ## Grade my continuity proof

Prove that if $f:R\rightarrow R$ is continuous and $f(x+y)=f(x)+f(y)$ for all $x,y\in R$, then there is a contstant $c\in R$, such that $f(x)=cx$, for all $x\in R$.

Here is my attempt:
Proof:
Let $f:R\rightarrow R$ and $f(x+y)=f(x)+f(y)$ for all $x,y\in R$, then there is a contstant $c\in R$, such that $f(x)=cx$, for all $x\in R$. Then, assume that f is NOT continuous. Then, suppose some point $a$ is not a limit point for $f$. Then, since $f(a)=ac$, $f$ is defined at $a$. Now,let $a$ be a limit point for $f$. Then, $0<|x-a|<\delta \Longrightarrow |f(x)-f(a)|<\epsilon$ for some $\epsilon>0$.
Then, $0<|x-a|<\delta \Longrightarrow |f(x)-f(a)|<\epsilon$ is the same as $0<|x-a|<\delta \Longrightarrow |cx-ca|<\epsilon$.
Now evaluate $|x-a|<\delta$. We have that $|x-a|<|c||x-a|=|cx-ca|<\epsilon$. Thus, $\delta =\epsilon$ works. However, we have shown that if $a$ is not a limit point at $f$ then $f$ is defined at $a$ and that if $a$ is a limit point of $f$ then $f(x)\longrightarrow f(a)$ as $x\longrightarrow a$. This is the definition of a continuous function so we have a contradiction. Thus, $f$ is continuous.

Any contructive critisism is welcome.

2. Hello, zebra.

Your proof does look somewhat confusing to me. I'll comment on parts of your proof.

Originally Posted by zebra2147
Prove that if $f:R\rightarrow R$ is continuous and $f(x+y)=f(x)+f(y)$ for all $x,y\in R$, then there is a contstant $c\in R$, such that $f(x)=cx$, for all $x\in R$.

Here is my attempt:
Proof:
Let $f:R\rightarrow R$ and $f(x+y)=f(x)+f(y)$ for all $x,y\in R$, then there is a contstant $c\in R$, such that $f(x)=cx$, for all $x\in R$.
What do you mean exactly by this? You seem to immediately conclude that there is a constant $c$, such that $f(x)=cx$ for all $x$, but this is what the whole problem is about provng.

Originally Posted by zebra2147
Then, assume that f is NOT continuous.
Why do you assume this? If you are going for either a proof by contraposition or a proof by contradiction, this is not the way to go.

Originally Posted by zebra2147
Then, suppose some point $a$ is not a limit point for $f$. Then, since $f(a)=ac$, $f$ is defined at $a$. Now,let $a$ be a limit point for $f$.
Why did you start out by assuming that $a$ is not a limit point of $f$, I mean, what do you use it for? You seem to conclude that $f$ is then defined at $a$, but $f$ is defined on all of the reals. And how do you know that $f(a) = ac$? Again it looks like you're using what you're actually supposed to prove.

I'm afraid that what you do is based on wrong assumptions, i.e. you assume what you want to end up proving.

3. Originally Posted by zebra2147
Prove that if $f:R\rightarrow R$ is continuous and $f(x+y)=f(x)+f(y)$ for all $x,y\in R$, then there is a contstant $c\in R$, such that $f(x)=cx$, for all $x\in R$.

Here is my attempt:
Proof:
Let $f:R\rightarrow R$ and $f(x+y)=f(x)+f(y)$ for all $x,y\in R$, then there is a contstant $c\in R$, such that $f(x)=cx$, for all $x\in R$.

Uh?? Why do you write the conclusion you must prove??

Then, assume that f is NOT continuous.

Why would you do that? This is a given, so you cannot assume it isn't true! You could negate the conclusion and try to prove by contradiction,

but this ain't it.

Then, suppose some point $a$ is not a limit point for $f$. Then, since $f(a)=ac$

This is wrong: you can't construct ANY prove assuming that what you HAVE to prove is correct!!

, $f$ is defined at $a$. Now,let $a$ be a limit point for $f$. Then, $0<|x-a|<\delta \Longrightarrow |f(x)-f(a)|<\epsilon$ for some $\epsilon>0$.
Then, $0<|x-a|<\delta \Longrightarrow |f(x)-f(a)|<\epsilon$ is the same as $0<|x-a|<\delta \Longrightarrow |cx-ca|<\epsilon$.
Now evaluate $|x-a|<\delta$. We have that $|x-a|<|c||x-a|=|cx-ca|<\epsilon$. Thus, $\delta =\epsilon$ works. However, we have shown that if $a$ is not a limit point at $f$ then $f$ is defined at $a$ and that if $a$ is a limit point of $f$ then $f(x)\longrightarrow f(a)$ as $x\longrightarrow a$. This is the definition of a continuous function so we have a contradiction. Thus, $f$ is continuous.

What a mess... ....this is NOT what you were asked to prove!!

You really must read much more carefully the problem and understand it, and then attempt to solve it.

Tonio

Any contructive critisism is welcome.
.

4. Hint for the OP:

Show that there exist c in R so that: f(x)=cx for all x in N(naturals), generalize it for all x in Z and then for all x in Q, conclude that f(x)=cx for all x in R. (Dense order)

5. Here is attempt #2:
Let's momentarily assume that the conclusion is correct.
Consider: $f(1) = 1c = c=f(1)$

Now, by induction, we can try to show that $f(xn)=f(x)*n$ for any positive integer n, and any real number,x. We showed the base case above. Now, the induction step...

$f(x * (n + 1)) = f(x * n) + f(x) = f(x) * n + f(x) = f(x) * (n + 1)$
This shows that if we have a positive integer n, then:

$f(n) = f(1 * n) = f(1) * n = cn$

Then, we show that it works for negatives...
$f(-x) + f(x) = f(-x + x) = f(0) = 0
$
So we have that, $f(-x) = -f(x)$

Now we try to show that $f(0)$ works too...
$f(0) = f(0 + 0) = f(0) + f(0)
$
. So we have that $f(0)=0$

We have shown that that for integers in $f$:
$f(n) = n * f(1)$

Using our results above we can look at rationals...
$f(p) = f(p/q * q)
= f(p/q) * q
f(p) / q = f(p/q)
f(p/q) = (p / q)c$

So we have shown that when $x$ is rational, $f(x)=cx$

Now, we have to show that this works for all real numbers...

First, suppose we have any real number x. Then, there exists a sequence of rationals, lets call it q(n), converging to that x. Above we showed that
$f(q(n)) = c * q(n) \longrightarrow cx$
However, since $f$ is continuous, this means that $cx = f(x)$. Therefore, for any $x$:
$f(x) = cx$

I think this is a better proof??? Although I know I kinda assumed the conclusion at the beginning. Any more feedback would be appreciated.

6. Originally Posted by zebra2147
Here is attempt #2:
Let's momentarily assume that the conclusion is correct.
Consider: $f(1) = 1c = c=f(1)$

Now, by induction, we can try to show that $f(xn)=f(x)*n$ for any positive integer n, and any real number,x. We showed the base case above. Now, the induction step...

$f(x * (n + 1)) = f(x * n) + f(x) = f(x) * n + f(x) = f(x) * (n + 1)$
This shows that if we have a positive integer n, then:

$f(n) = f(1 * n) = f(1) * n = cn$

Then, we show that it works for negatives...
$f(-x) + f(x) = f(-x + x) = f(0) = 0
$
So we have that, $f(-x) = -f(x)$

Now we try to show that $f(0)$ works too...
$f(0) = f(0 + 0) = f(0) + f(0)
$
. So we have that $f(0)=0$

We have shown that that for integers in $f$:
$f(n) = n * f(1)$

Using our results above we can look at rationals...
$f(p) = f(p/q * q)
= f(p/q) * q
f(p) / q = f(p/q)
f(p/q) = (p / q)c$

So we have shown that when $x$ is rational, $f(x)=cx$

Now, we have to show that this works for all real numbers...

First, suppose we have any real number x. Then, there exists a sequence of rationals, lets call it q(n), converging to that x. Above we showed that
$f(q(n)) = c * q(n) \longrightarrow cx$
However, since $f$ is continuous, this means that $cx = f(x)$. Therefore, for any $x$:
$f(x) = cx$

I think this is a better proof??? Although I know I kinda assumed the conclusion at the beginning. Any more feedback would be appreciated.

Now the above almost begins to make sense. To begin with, $f(0)=f(0+0)=f(0)+f(0)\Longrightarrow f(0)=0$.

Now, if $n\in\mathbb{N}$ , then $f(n)=f(1+...+1)=nf(1)$ (this can be shown this by induction, as you did), so putting $f(1):=c$ we get that $\forall n\in\mathbb{N}\,,\,\,f(n)=cn$.

Now take it from here just polishing up what you already did above.

Tonio