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limit of f(x+y)=f(x)+f(y) near 0

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Let $f: \mathbb{R}\to \mathbb{R}$ be such that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ . Assume the limit of $\lim_{x\to 0}f(x)=L$ then prove that $L=0$.

Here is my attempt:

We first choose an arbitrary \(\displaystyle \epsilon >0\) then we choose $\delta$ such that \(\displaystyle \text{Max}(|x|,|y|,|x+y|)<\delta \) so we can say that \(\displaystyle |f(y)-L|<\frac{\epsilon}{2} \) and \(\displaystyle |f(x+y)-L|<\frac{\epsilon}{2} \)

Since we have the following

\(\displaystyle |f(x+y)-L|=|f(x)+f(y)-L| \geq |f(x)|-|f(y)-L|\)

\(\displaystyle |f(x)|\leq |f(x+y)-L|+|f(y)-L|<\frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon \) for the chosen \(\displaystyle \delta \) by existence of limit of \(\displaystyle f\).

So we deduce that

\(\displaystyle 0<|x|<\delta \) implies \(\displaystyle |f(x)|<\epsilon \)

since \(\displaystyle \epsilon >0\) is arbitrary we have \(\displaystyle \lim_{x \to 0}f(x)=0 \,\,\, \square \).

So what you think guys ?
 

zzephod

Well-known member
Feb 3, 2013
134
Let $f: \mathbb{R}\to \mathbb{R}$ be such that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ . Assume the limit of $\lim_{x\to 0}f(x)=L$ then prove that $L=0$.

Here is my attempt:

We first choose an arbitrary \(\displaystyle \epsilon >0\) then we choose $\delta$ such that \(\displaystyle \text{Max}(|x|,|y|,|x+y|)<\delta \) so we can say that \(\displaystyle |f(y)-L|<\frac{\epsilon}{2} \) and \(\displaystyle |f(x+y)-L|<\frac{\epsilon}{2} \)

Since we have the following

\(\displaystyle |f(x+y)-L|=|f(x)+f(y)-L| \geq |f(x)|-|f(y)-L|\)

\(\displaystyle |f(x)|\leq |f(x+y)-L|+|f(y)-L|<\frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon \) for the chosen \(\displaystyle \delta \) by existence of limit of \(\displaystyle f\).

So we deduce that

\(\displaystyle 0<|x|<\delta \) implies \(\displaystyle |f(x)|<\epsilon \)

since \(\displaystyle \epsilon >0\) is arbitrary we have \(\displaystyle \lim_{x \to 0}f(x)=0 \,\,\, \square \).

So what you think guys ?

Observe that: \(f(x)=2f(x/2)\), then as \(\lim_{x \to 0} f(x) = \lim_{x \to 0} f(x/2)\) we have \(L=2L\) etc

.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725

Pranav

Well-known member
Nov 4, 2013
428
Not a rigorous proof but f(x)=kx (k is a constant) satisfies the given functional relation so the limit is zero.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Not a rigorous proof but f(x)=kx (k is a constant) satisfies the given functional relation so the limit is zero.
I don't get your point .
 

Pranav

Well-known member
Nov 4, 2013
428
I don't get your point .
I am sorry if I used the wrong words. I have no knowledge of those epsilon and delta proofs you deal with. The functional relation "f(x+y)=f(x)+f(y)" is a common one for me (I mean I see it frequently while doing problems). The function f(x)=kx satisfies the above relation i.e f(x+y)=kx+ky and f(x)+f(y)=kx+ky.
So $\lim_{x \rightarrow 0} kx=0$. You may ignore my post if you wish to. :)
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
I am sorry if I used the wrong words. I have no knowledge of those epsilon and delta proofs you deal with. The functional relation "f(x+y)=f(x)+f(y)" is a common one for me (I mean I see it frequently while doing problems). The function f(x)=kx satisfies the above relation i.e f(x+y)=kx+ky and f(x)+f(y)=kx+ky.
So $\lim_{x \rightarrow 0} kx=0$. You may ignore my post if you wish to. :)
[JUSTIFY]I think what Zaid means is that while you showed that this particular function works, that does not really show that all functions that satisfy $f(x + y) = f(x) + f(y)$ meet the given condition (though in this case they happen to). As an example: just because you are an internet user, and also an MHB member, doesn't mean all internet users are also MHB members.[/JUSTIFY]
 

Pranav

Well-known member
Nov 4, 2013
428
[JUSTIFY]I think what Zaid means is that while you showed that this particular function works, that does not really show that all functions that satisfy $f(x + y) = f(x) + f(y)$ meet the given condition (though in this case they happen to). As an example: just because you are a internet user, and also an MHB member, doesn't mean all internet users are also MHB members.[/JUSTIFY]
Thanks Bacterius!

I am interested to know if there are any other functions which satisfy the given functional relation. Do you have any ideas? :)
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Thanks Bacterius!

I am interested to know if there are any other functions which satisfy the given functional relation. Do you have any ideas? :)
Hmm. There are discontinuous functions which satisfy the relation (see Cauchy's functional equation - Wikipedia, the free encyclopedia) but Zaid's claim requires continuity at $x = 0$ for the limit to exist. So they may or may not apply to the problem, I am not sure. In any case, it is still important to verify that the claim still holds over (or applies to) all such functions, otherwise the proof does not quite follow.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Thanks Bacterius!

I am interested to know if there are any other functions which satisfy the given functional relation. Do you have any ideas? :)
How about:
$$f(x) = \begin{cases}
3x & \text{ if } x \in \mathbb Q \\
-2x & \text{ otherwise }
\end{cases}$$
 

Pranav

Well-known member
Nov 4, 2013
428
How about:
$$f(x) = \begin{cases}
3x & \text{ if } x \in \mathbb Q \\
-2x & \text{ otherwise }
\end{cases}$$
Sorry for the dumb question, how do I evaluate f(x+y), f(x) and f(y) here? :confused:
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Define the Laplace transform as follows

\(\displaystyle F(s)=\mathcal{L}(f(t))=\int^{\infty}_0 e^{-st}\, f(t)\, dt \)

Take the space of all functions that have a Laplace transform then

\(\displaystyle \mathcal{L}(x(t)+y(t))=\int^{\infty}_0 e^{-st}\, (x(t)+y(t))\, dt = \int^{\infty}_0 e^{-st}\, x(t)\, dt \, +\,\int^{\infty}_0 e^{-st}\, y(t)\, dt \,= \mathcal{L}(x(t))+\mathcal{L}(y(t)) \)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
How about:
$$f(x) = \begin{cases}
3x & \text{ if } x \in \mathbb Q \\
-2x & \text{ otherwise }
\end{cases}$$
Sorry for the dumb question, how do I evaluate f(x+y), f(x) and f(y) here? :confused:
Sorry, I have just realized that my suggestion does not work.
$$f(1+\pi) = -2(1+\pi) = -2 -2\pi$$
$$f(1) + f(\pi) = 3 \cdot 1 + (-2)\cdot \pi = 3 - 2\pi$$
Obviously they are not the same.

EDIT: The following function does work though.
$$f(x) = \begin{cases}
0 & \text{ if } x \in \mathbb Q \\
2x & \text{ otherwise }
\end{cases}$$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Define the Laplace transform as follows

\(\displaystyle f(s)=\mathcal{L}(f(t))=\int^{\infty}_0 e^{-st}\, f(t)\, dt \)

Take the space of all functions that have a Laplace transform then

\(\displaystyle \mathcal{L}(x(t)+y(t))=\int^{\infty}_0 e^{-st}\, (x(t)+y(t))\, dt = \int^{\infty}_0 e^{-st}\, x(t)\, dt \, +\,\int^{\infty}_0 e^{-st}\, y(t)\, dt \,= \mathcal{L}(x(t))+\mathcal{L}(y(t)) \)
Hmm, but $\mathcal{L}$ isn't a function of $\mathbb R \to \mathbb R$ is it?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
We can have a generalization for all linear operators

\(\displaystyle L(f+g)=L(f)+L(g) \)

\(\displaystyle L(k\, f) = k\, L(f) \,\,\,\, \) where k is constant
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Let $f: \mathbb{R}\to \mathbb{R}$ be such that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ . Assume the limit of $\lim_{x\to 0}f(x)=L$ then prove that $L=0$.
Another way: $L=\displaystyle\lim_{x\to 0}f(x)$ implies $\displaystyle\lim_{n\to +\infty}f\left(\frac{1}{n}\right)=L.$

But (easily proved), $\displaystyle f\left(\frac{1}{n}\right)=\frac{f(1)}{n},$ so $L=\displaystyle\lim_{n\to +\infty}\frac{f(1)}{n}=0.$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Hmm, but $\mathcal{L}$ isn't a function of $\mathbb R \to \mathbb R$ is it?
Take the space of all constant functions in \(\displaystyle \mathbb{R}\) then \(\displaystyle x(t) =a , y(t) = b \)

\(\displaystyle \mathcal{L}(a+b)=\mathcal{L}(a)+\mathcal{L}(b)\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Take the space of all constant functions in \(\displaystyle \mathbb{R}\) then \(\displaystyle x(t) =a , y(t) = b \)

\(\displaystyle \mathcal{L}(a+b)=\mathcal{L}(a)+\mathcal{L}(b)\)
Ah, but when you write $\mathcal{L}(a)$, with $a$ you mean a constant function, which is an element of $\mathbb R \to \mathbb R$ given by $t \mapsto a$, which is still not an element of $\mathbb R$.

More importantly, $\mathcal{L}(a)$ is also a function given by \(\displaystyle s \mapsto \frac a s\), which is not an element of $\mathbb R$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
EDIT: The following function does work though.
$$f(x) = \begin{cases}
0 & \text{ if } x \in \mathbb Q \\
2x & \text{ otherwise }
\end{cases}$$
Sorry, that doesn't work either. For example, if $x=\pi$ and $y=4-\pi$ then $f(x)+f(y) = 2\pi + 2(4-\pi) = 8$ but $f(x+y) = f(4) = 0.$

To construct a discontinuous function on $\mathbb{R}$ that satisfies $f(x+y) = f(x) + f(y)$, you need to define it on the elements of a Hamel basis (see the link in Bacterius's comment #9). The existence of a Hamel basis depends in turn on the axiom of choice. I am fairly sure that it is not possible to find an example of a discontinuous $f$ without using the axiom of choice.