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Let $f:\mathbb{R} \to \mathbb{R}$ be a function that's continuous at the origin. Show that if $$f(x+y)=f(x)+f(y)$$ for all $x, y \in\mathbb{R}$, then $f$ is continuous.

This was a problem on our homework and the TA's suggestion for the solution was the following:

Let's first show that $f(0) = 0$. Since $f(0) =f(0+0) = f(0)+f(0)$

we have that $f(0)=0$.

Now we get

$$\lim_{y\to x} f(y)=\lim_{h\to 0} f(x+h) = \lim_{h\to 0} (f(x) +f(h)) = \lim_{h\to 0} f(x) + \lim_{h\to 0} f(h) = f(x) +0=f(x).$$

I'm not very satisfied with this. There's no explanation for the result at all. Could someone open this up for me?

  1. How can it be for sure that $f(0) =0$? They haven't given us any $f$. For all I know $f$ could be $f(x)= x^{73}+1$ and $f(0) = 1$?

  2. Where does he come up with $\lim_{y\to x}f(y)$ and how come this is equal to $\lim_{h\to 0} f(x+h)$?

  3. How does he get the result $\lim_{h\to 0} f(x) + \lim_{h\to 0} f(h) = f(x) +0$?

  • If $f(0)=1$ then $f(0)=f(0+0)=f(0)+f(0)=2f(0)=2 \times 1=2 \neq 1=f(0)$ – For the love of maths Apr 09 '20 at 20:10
  • What do you mean? You've just proven that $f(0) = 0$ from the fact that $f(x+y) = f(x) + f(y)$ for all $x,y$. You know that you can't have $f(x) = x^{73} + 1$ because it doesn't obey that very special property. – Jair Taylor Apr 09 '20 at 20:14
  • Subtract $f(0)$ from both sides of $f(0) = f(0) + f(0)$ and you see that $f(0) = 0$. Maybe that is the step you missed? – Jair Taylor Apr 09 '20 at 20:16
  • Aczel, 1969, "On Applications and Theory of Functional Equations" –  Apr 09 '20 at 20:28

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