Locally Bounded Functional Equation $f(x+y) = f(x) + f(y)$ and Continuity 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!
You can show that $f(0)=0$ using the additivity property of $f$.  

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

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

Then $Nx_\delta$ is in $(-c,c)$ and
$$|f( N {x_\delta })| =N |f(x_\delta )|&gt;N\epsilon&gt;M,$$ 
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 $\alpha$).
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