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

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In summary: 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... hold water.
  • #1
alyafey22
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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 ?
 
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  • #2
ZaidAlyafey said:
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

.
 
  • #3
ZaidAlyafey said:
So what you think guys ?
Your proof looks fine (though it's not as slick as zzephod's).
 
  • #4
Not a rigorous proof but f(x)=kx (k is a constant) satisfies the given functional relation so the limit is zero.
 
  • #5
Pranav said:
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 .
 
  • #6
ZaidAlyafey said:
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. :)
 
  • #7
Pranav said:
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. :)

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.​
 
  • #8
Bacterius said:
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.​

Thanks Bacterius!

I am interested to know if there are any other functions which satisfy the given functional relation. Do you have any ideas? :)
 
  • #9
Pranav said:
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.
 
  • #10
Pranav said:
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:
$$f(x) = \begin{cases}
3x & \text{ if } x \in \mathbb Q \\
-2x & \text{ otherwise }
\end{cases}$$
 
  • #11
I like Serena said:
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:
 
  • #12
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:
  • #13
I like Serena said:
How about:
$$f(x) = \begin{cases}
3x & \text{ if } x \in \mathbb Q \\
-2x & \text{ otherwise }
\end{cases}$$

Pranav said:
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.
$$f(x) = \begin{cases}
0 & \text{ if } x \in \mathbb Q \\
2x & \text{ otherwise }
\end{cases}$$
 
  • #14
ZaidAlyafey said:
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?
 
  • #15
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
 
  • #16
ZaidAlyafey said:
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 $L=\displaystyle\lim_{n\to +\infty}\frac{f(1)}{n}=0.$
 
  • #17
I like Serena said:
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 \(\displaystyle x(t) =a , y(t) = b \)

\(\displaystyle \mathcal{L}(a+b)=\mathcal{L}(a)+\mathcal{L}(b)\)
 
  • #18
ZaidAlyafey said:
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 $\mathbb R$.
 
  • #19
I like Serena said:
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 http://mathhelpboards.com/analysis-50/limit-f-x-y-%3Df-x-f-y-near-0-a-7582.html#post34576). 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.
 

FAQ: Limit of f(x+y)=f(x)+f(y) near 0

What is the meaning of the limit of f(x+y) as x and y approach 0?

The limit of f(x+y) as x and y approach 0 means the value that f(x+y) approaches as x and y get closer and closer to 0. In other words, it is the value that f(x+y) gets infinitely close to without ever actually reaching 0.

How do you calculate the limit of f(x+y) near 0?

To calculate the limit of f(x+y) near 0, you can use the definition of a limit. This involves taking the limit as x and y approach 0 of the expression f(x+y) and evaluating it. Alternatively, you can also use algebraic techniques such as factoring or simplifying the expression to evaluate the limit.

Why is the limit of f(x+y) near 0 important?

The limit of f(x+y) near 0 is important because it helps us understand the behavior of a function as the input values get closer and closer to 0. It can also help us determine the continuity of a function and identify any potential discontinuities or singularities.

Can the limit of f(x+y) near 0 be different from the value of f(0)?

Yes, the limit of f(x+y) near 0 can be different from the value of f(0). This can happen if the function is not defined at x=0 or if there is a discontinuity or singularity at x=0. In these cases, the limit of f(x+y) near 0 may not exist or may be different from the value of f(0).

How does the limit of f(x+y) near 0 relate to the concept of continuity?

The limit of f(x+y) near 0 is closely related to the concept of continuity. A function is considered continuous at a point if the limit of the function at that point exists and is equal to the value of the function at that point. Therefore, if the limit of f(x+y) near 0 exists and is equal to f(0), the function is continuous at x=0.

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