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Let $f$ be a real-valued function on $\mathbb{R}$ s.t. $f(x+y) = f(x) + f(y)$ for all $x,y$ reals. Suppose there are reals $c$ and $M$ s.t. $|f(x)| \leq M $ for all $x$ in $[-c,c]$. Show that $f$ is continuous. I am able to show that $f$ must take the form $f(x) = xf(1)$ for $x$ rational, but am having trouble showing this holds for the irrationals as well. Hints appreciated!

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Hint: f(x-y) = f(x)-f(y), so if you find two numbers x and y such that they are less than c apart but f(x) and f(y) are more than M apart, you have a contradiction. Further hint: use the fact that you can "blow the function up" using f(cx)=cf(x) for any integer c. With this, you can translate any small jump in the function into a really big one later down the line. –  Lopsy Feb 10 '12 at 22:03
Duplicate? –  JavaMan Feb 10 '12 at 22:03
I don't think that's quite a duplicate, but that question and en.wikipedia.org/wiki/Cauchy%27s_functional_equation are both pretty good references for this problem. –  Lopsy Feb 10 '12 at 22:06
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3 Answers

up vote 4 down vote accepted

You can show that $f(0)=0$ using the additivity property of $f$.

Suppose $f$ is not continuous at $0$. Then there is an $\epsilon>0$ so that for any $\delta>0$, there is an $x_\delta$ satisfying both $|x_\delta|<\delta$ and $|f(x_\delta)|>\epsilon$.

Let $N$ be a positive integer so that $N\epsilon>M$ and take $\delta=c/N$. Now choose $x_\delta$ satisfying both $|x_\delta|<\delta$ and $|f(x_\delta)|>\epsilon$.

Then $Nx_\delta$ is in $(-c,c)$ and $$|f( N {x_\delta })| =N |f(x_\delta )|>N\epsilon>M,$$ a contradiction.

Thus, $f$ is continuous at $0$; and, by the additivity of $f$, on $\Bbb R$ as well.

Indeed, this is easily proven using sequences: Let $y\in\Bbb R$ and suppose $y_n\rightarrow y$. Then $y_n-y\rightarrow 0$. Since $f$ is continuous at $0$, it follows that $f(y_n-y)$ converges to 0. By the graces of the additivity of $f$, it follows that $f(y_n)$ converges to $f(y)$; whence, $f$ is continuous at $y$.

(Note, now with continuity in hand, you can obtain that $f(\alpha x)=\alpha f(x)$ for all $\alpha$ by considering rationals $q_\alpha$ converging to $\alpha$).

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In fact, following this technique, a stronger result is that $f$ is uniformly continuous on $\mathbb{R}$. If not, we choose $|x-y| < \delta$ but $|f(x-y)| > \epsilon$. The proof precedes as the first part, and avoids dealing with the special case zero. –  Peter Feb 12 '12 at 6:04
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$$f(0+x)=f(0)+f(x)=f(x) \implies f(0)=0$$

$$\forall y\in\mathbb{R}:\lim_{\epsilon \to \pm 0}f(y+\epsilon)=\lim_{\epsilon \to \pm0}f(y)+f(\epsilon)=f(y)+\lim_{\epsilon \to \pm0}{f(\epsilon)}=f(y)$$

this last line proves continuity

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You assumed that $\lim_{\epsilon\to 0} f(\epsilon) = 0$. Also, he wants continuity everywhere, not just $[-c,c]$ –  Thomas Andrews Feb 10 '12 at 22:16
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Use the definition of continuity:

$\frac{f(x+\epsilon)-f(x)}{ \epsilon}=\frac{f(\epsilon)}{\epsilon}\leq \frac{M}{\epsilon} \forall \epsilon \in [-c,c]$

Update: $\forall \epsilon \in [-c,c]-\{0\}$

Update 2: here is the definition: $f$ is continuous at $x$ iff $\lim_{\epsilon\to x} \frac{f(x+\epsilon)-f(x)}{ \epsilon} $ is defined.

Update 3: all above is about the differentiability and not continuity; and its a working proof too

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Huh, that's not the definition of continuity in any book that I've ever seen. The LHS of your formula should be $\frac{f(\epsilon)}{\epsilon}$, not $f(\epsilon)$. –  Thomas Andrews Feb 10 '12 at 22:18
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