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I'm new here and could really use some help please:

Let $f$ be an additive function. So for all $x,y \in \mathbb{R}$, $f(x+y) = f(x)+f(y)$.

  1. Prove that if there are $M>0$ and $a>0$ such that if $x \in [-a,a]$, then $|f(x)|\leq M$, then $f$ has a limit at every $x\in \mathbb{R}$ and $\lim_{t\rightarrow x} f(t) = f(x)$.

  2. Prove that if $f$ has a limit at each $x\in \mathbb{R}$, then there are $M>0$ and $a>0$ such that if $x\in [-a,a]$, then $|f(x)| \leq M$.

if necessary the proofs should involve the $\delta - \varepsilon$ definition of a limit.


The problem had two previous portions to it that I already know how to do. However, you can reference them to do the posted portions of the problem. Here they are:

(a) Show that for each positive integer $n$ and each real number $x$, $f(nx)=nf(x)$.

(b) Suppose $f$ is such that there are $M>0$ and $a>0$ such that if $x\in [−a,a]$, then $|f(x)|\le M$. Choose $\varepsilon > 0$. There is a positive integer $N$ such that $M/N < \varepsilon$. Show that if $|x-y|<a/N$, then |$f(x)-f(y)|<\varepsilon$.

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  • $\begingroup$ The problem had two previous portions to it that i already know how to do. However, you can reference them to do the posted portions of the problem. Here they are: (a) show that for each positive integer n and each real number x, f(nx)=nf(x). (b) suppose f is such that there are M>0 and a>0 such that if $x \in [-a,a]$, then $|f(x)| \leq M$. Choose epsilon > 0. There is a positive integer N such that M/N < epsilon. Show that if |x-y|<a/N, then |f(x)-f(y)|<epsilon. $\endgroup$ – user97065 Sep 26 '13 at 12:29
  • $\begingroup$ user97065: I've added your previous comment to your question. I consider it part of background for this question. It is better to provide the context directly in your post (where it is much more prominently visible) than in comments (which can be easily missed by a reader). $\endgroup$ – Martin Sleziak Sep 26 '13 at 14:55
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Your Question is completely answered in the following book (Theorem 1.2, page 3). Also you can find some further related topics.

Pl. Kannappan, Functional Equations and Inequalities with Applications, Springer, 2009.

Theorem 1.2. Suppose $A : R\rightarrow R$ satisfies $A(x+y)=A(x)+A(y)$ with $c = A(1) > 0$. Then the following conditions are equivalent:

(i) $A$ is continuous at a point $x_{0}$.

(ii) $A$ is monotonically increasing.

(iii) $A$ is nonnegative for nonnegative $x$.

(iv) $A$ is bounded above on a finite interval.

(v) $A$ is bounded below on a finite interval.

(vi) $A$ is bounded above (below) on a bounded set of positive Lebesgue measure.

(vii) $A$ is bounded on a bounded set of positive measure (Lebesgue).

(viii) $A$ is bounded on a finite interval.

(ix) $A(x) = cx$.

(x) $A$ is locally Lebesgue integrable.

(xi) $A$ is differentiable.

(xii) A is Lebesgue measurable.

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Hints: 1. For any $\varepsilon>0$ let $\delta := \frac{a}{N}$ and apply (b).
2. It is obvious that $f(0) = 0$. Since $f$ has a limit at $0$, we can put $\varepsilon = 1$ and write: $\exists \delta > 0,\ \forall x,\ |x|<\delta\colon\ |f(x)|<1$.

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  • $\begingroup$ Thank you soooo much for your help! i have a question though for #1: how would one generalize the delta-epsilon definition for every x in R, not for just a single x? $\endgroup$ – user97065 Sep 26 '13 at 16:54
  • $\begingroup$ The delta-epsilon definition of what? $\endgroup$ – njguliyev Sep 26 '13 at 19:09
  • $\begingroup$ the delta-epsilon definition of a limit $\endgroup$ – user97065 Sep 26 '13 at 22:34

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