# Functions and limits

I find it difficult to solve problems of this type. I do not understand how to start and then develop the proof. For example, I tried to solve this question:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function such that for $M>0$, satisfies: $$|f(x_1+x_2+x_3+\cdot+x_n)-f(x_1)-f(x_2)-\cdots-f(x_n)|\le M$$ $\forall x_1,x_2,\cdots,x_n \in \mathbb{R}.$

Prove that:

$$f(x+y)=f(x)+f(y),\qquad \forall x,y \in\mathbb{R}$$

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As to approach, experience with functional equations in "contest math" helps. One would almost automatically first look at $f(0)$, then $f(-x)$. Then things click into place. – André Nicolas Jan 2 '12 at 21:39

I'll try and give some insight into how I'd think my way through this problem.

You're told that the difference between the function applied to a sum of $n$ numbers and the sum of the function applied to those numbers is bounded by some constant. Since we have freedom over our choice of $n$, chances are we're going to get something like $\frac{M}{n}$ to play with, because this might provide us with a bound tending to zero.

So let's work backwards. We're looking for something like $|f(x+y) - f(x) - f(y)| \le \frac{k}{n}$ for some constant $k$.

Multiplying through by $n$ we get $|nf(x+y)-nf(x)-nf(y)| \le k$. It's starting to look more familiar now, but it's not in quite the right form. We can get $nf(x+y)$ from by considering $f(n(x+y))$ and $nf(x)+nf(y)$ by considering $f(nx+ny)$... but wait, these are the same thing! So if we add and subtract then we might be able to use the triangle inequality cunningly.

And we can:

$|nf(x+y) - nf(x) - nf(y)| \le |nf(x+y) - f(n(x+y))| + |f(n(x+y)) - nf(x) - nf(y)|$

That $|f(n(x+y)) - nf(x+y)| \le M$ and $|f(n(x+y)) - nf(x) - nf(y)| \le M$ follow immediately from the definition, in the former case by considering $n$ lots of $x+y$ and in the second by considering $n$ lots of $x$ and $n$ lots of $y$.

So we can pick $k=2M$ and working in reverse we arrive at our conclusion that, for any $n$, $|f(x+y)-f(x)-f(y)| \le \frac{2M}{n}$, and the result follows.

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I interpret the question as follows:

Suppose there exists $f : \mathbb R \to \mathbb R$ and $M \geqslant 0$ such that for all $n$ and all $n$-tuples $(x_1, \ldots, x_n)$ we have $$|f(x_1 + \cdots x_n) - f(x_1) - \cdots - f(x_n)| \leqslant M. \tag{\ast}$$ Show that $f (x+y) = f(x) + f(y)$.

Fix $x, y$.

1. Taking $n$ copies of $x$ followed by $n$ copies of $y$ in $(\ast)$, we obtain $$|f(nx + ny) - n f(x) - nf(y)| \leqslant M.$$

2. Replace $y$ by $0$ and $x$ by $x+y$ in $(1)$ to get $$|f(nx+ny) - n f(x+y) - n f(0)| \leqslant M.$$

3. Indeed, we can show* that $f(0) = 0$ so that $$|f(nx+ny) - n f(x+y)| \leqslant M.$$

4. From $(1)$ and $(3)$ and the triangle inequality, we have \begin{align*} |nf(x+y) - nf(x) - nf(y)| &= | \left( nf(x+y) - f(nx+ny) \right) + \left( f(nx+ny) - nf(x) - nf(y) \right)| \\&\leqslant M + M = 2M. \end{align*} That is, we have $|f(x+y) - f(x) - f(y) | \leqslant \frac{2M}{n}$ for all $n$. By the Archimedean property of the reals, we can conclude that $f(x+y) = f(x) + f(y)$.

*To show that $f(0) = 0$, take $n$ copies of $0$ in $(\ast)$: $$(n-1) \cdot |f(0)| = |f(0) - n f(0)| \leqslant M,$$ which is possible only if $f(0) = 0$. Thanks to Didier for pointing out this gap.

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I guess that a zeroth step is omitted, which is to show that $f(0)=0$. – Did Jan 2 '12 at 20:48
@Didier, Oops, more like forgotten due to carelessness. :-) I will edit now. Thanks! – Srivatsan Jan 2 '12 at 20:50

You can use the Archimedean property to prove in turn

1. $f(0)=0$
2. $f(-x)=-f(x)$
3. $f(x+y)=f(x)+f(y)$

by looking at

$$|f(0+0+0+\cdots+0)-f(0)-f(0)-\cdots-f(0)|\le M$$

$$|f(x-x+x+\cdots-x)-f(x)-f(-x)-f(x)-\cdots-f(-x)|\le M$$

$$|f(x+y+(-x-y)+\cdots+x+y+(-x-y))-f(x)-f(y)-f(-x-y)-\cdots-f(x)-f(y)-f(-x-y)|\le M$$

For example, if $f(0)\not = 0$ then there is an $n$ large enough that the first expression exceeds a given $M$.

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