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If $f(x)+f(y)\leq f(x+y)$ and $f:\mathbb{R}\to\mathbb{R}$, then can we find $\lim_{x\to 0} \frac {f(x)}{x}$?

I am not sure whether the question is correct.Thank you.(I tried this idea: $f(x)=f(x+y-y)\ge f(x+y)+f(-y)\ge f(x)+f(y)+f(-y)\implies f(y)\leq -f(-y)$ but after that I seem to be hitting a roadblock.

Thank you in advance.

Martin Sleziak
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Eisen
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    Are you assuming continuity or anything? There exist discontinuous $f$ such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb R$, in for those, the limit will not exist. – Jonas Meyer Jan 21 '12 at 06:13
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    To add to @Jonas' remark: Every function of the form $f(x) = ax$ (with $a$ real) satisfies the inequation. For such a function, the limit exists and is equal to $a$. So it is possible that the limit exists but cannot be found from the given data alone. Can you clarify what you really want to show? – Srivatsan Jan 21 '12 at 06:18
  • @Sri : In your example the limit is $f'(x)$ for any $x \in \mathbb{R}$...it is not conclusive – Rajesh D Jan 21 '12 at 06:25
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    @Rajesh: Srivatsan's point is that even among functions for which the limit exists, it is not uniquely determined from the hypotheses. There exist $f$ where the limit is $0$, $46$, or $-\sqrt 2$, and others for which it doesn't exist. – Jonas Meyer Jan 21 '12 at 06:35
  • @Jonas : I got it, i didn't see it along side with your comment. Could you give any examples of the type of functions you have mentioned ? – Rajesh D Jan 21 '12 at 06:47
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    @Jonas: However, I think it pretty clear that the OP wants to ask whether the hypothesis implies that the limit exists, not whether the value of the limit can be inferred from the hypothesis alone. – Brian M. Scott Jan 21 '12 at 06:48
  • @Rajesh: The functions that Srivatsan mentioned will do. – Brian M. Scott Jan 21 '12 at 06:49
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    @Brian: Your guess may well be right, but it is not clear to me. "Can we find" the limit often means more, namely finding the value of the limit. Hopefully Sabyasachi Mukherjee will clarify. – Jonas Meyer Jan 21 '12 at 07:29
  • Analogous question for $x\to\infty$ is answered here: Fekete's lemma for real functions. Although I am not sure that would be of much help for $x\to0$. – Martin Sleziak Jan 21 '12 at 10:17
  • I didn't quite think of that(other conditions).Either way,thanks for your comments. – Eisen Jan 21 '12 at 14:16

1 Answers1

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A function $f:{\mathbb R}\to{\mathbb R}$ which fulfills $$(\forall x,y\in\mathbb R)f(x+y) \le f(x)+f(y)$$ is called is subadditive.

It was already mentioned in comments that the limit need not exist without any additional assumptions on $f$. E.g. if $f$ is any non-linear solution of Cauchy's equation, then the limits does not exists, but the function is both subadditive and superadditive.

The following result from the book An introduction to the theory of functional equations and inequalities By Marek Kuczma p.467 gives at least some conditions when the limit exists:

Theorem 16.3.3. Let $f:\mathbb R\to\mathbb R$ be a measurable subadditive function, and let $$A = \inf_{t<0} \frac{f(t)}t, \qquad B=\sup_{t>0} \frac{f(t)}t.$$ If $A$ resp. $B$ is finite, then $$A = \lim_{h\to0^-} \frac {f(h)}h,\text{ resp. }B=\lim_{h\to0^+} \frac {f(h)}h.$$ The above formulas remain valid for $A$ and/or $B$ infinite under the additional assumption that $\lim\limits_{x\to 0} f(x) = 0$, or $\liminf\limits_{x\to 0} f(x)>0$ Moreover, in every case, $$A \le B.$$

If you rewrite the above results for the function $g(x)=-f(x)$, you get results for superadditive functions, i.e. $g(x+y)\ge g(x)+g(y)$.

Martin Sleziak
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  • To be honest, I have not yet completed a first course in rigorous calculus(I am currently doing Apostol' Calculus vol I and I haven't done much of it(only hundred odd pages).So I don't think I have the necessary background to read Kuzma's book(I can't see the preview of that page either).But still thank you.I don't even know what a measurable function is.So I really apologise for my ignorance. – Eisen Jan 21 '12 at 14:14