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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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
So what you think guys ?
Your proof looks fine (though it's not as slick as zzephod's).

#### Pranav

##### Well-known member
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
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
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
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
[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
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
Thanks Bacterius!

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

#### Pranav

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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

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

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
$$f(x) = \begin{cases} 3x & \text{ if } x \in \mathbb Q \\ -2x & \text{ otherwise } \end{cases}$$
Sorry for the dumb question, how do I evaluate f(x+y), f(x) and f(y) here?
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
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
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
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
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
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
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.