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Let $x,y \in R$

$f(x+y) = f(x)+f(y)$

is it true that if $f$ is continuous at $x_0=0$, than $f$ is continuous in $R$?

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closed as off-topic by zhoraster, user223391, Namaste, Did, Adam Hughes Dec 18 '16 at 2:02

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    $\begingroup$ Hint: fix a $t\in\mathbb R$ and consider the function $g(x)=f(x+t)$. $\endgroup$ – Wojowu Dec 17 '16 at 21:43
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At any arbitrary $x_1\in\mathbb{R}$ and any $\Delta\neq 0$, we have $$ f(x_1+\Delta)-f(x_1)=f(\Delta)=f(\Delta+x_0)-f(x_0). $$ As $\Delta\to 0$, the rightmost expression above goes to $0$ due to continuity at $x_0$, so the leftmost expression also goes to $0$. This implies continuity at $x_1$ and therefore in $\mathbb{R}$. Note we don't need the fact that $x_0=0$.

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    $\begingroup$ Wondrously simple. +1 $\endgroup$ – Simply Beautiful Art Dec 17 '16 at 22:06
  • $\begingroup$ I don't understand how and why you used the continuity at $x_0$. Couldn't you say immediately that $f(x_1+\Delta)-f(x_1)$ goes to zero when $\Delta$ goes to zero because it can be seen as $f(x_1 + 0)-f(x_1) = 0$? $\endgroup$ – S. Peter Dec 17 '16 at 22:20
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    $\begingroup$ @S.Peter $\lim_{\Delta\to 0}f(x_1+\Delta)=f(x_1)$ is what is to be proved for all $x_1\in\mathbb{R}$. However, you're given $\lim_{\Delta\to 0}f(x_0+\Delta)=f(x_0)$, so I used this to infer that $\lim_{\Delta\to 0}[f(\Delta+x_0)-f(x_0)]=0$. The key is to note that $f(x+y)=f(x)+f(y)$ implies $f(\Delta+x_0)-f(x_0)=f(\Delta+x_1)-f(x_1) $. $\endgroup$ – yurnero Dec 17 '16 at 22:25
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    $\begingroup$ @S.Peter $\lim_{x\to x_1}f(x)=\lim_{\Delta\to 0}f(x_1+\Delta)$. $\Delta$ represents $x-x_1$. $\endgroup$ – yurnero Dec 17 '16 at 22:54
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Make $x=a_n$ such that $a_n \rightarrow a$ and $y=-a$

$$f\left(a_n-a\right)=f(a_n)-f\left(a\right)$$

doing $n \rightarrow \infty$ and by the continuity at $0$

$$f\left(a_n-a\right) \rightarrow 0 \quad \text{(because $f(0)=0$)}$$

$$a_n \rightarrow a$$

and finally

$$f\left(a_n\right) \rightarrow f(a)$$

So we get continuity at $a$.

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  • $\begingroup$ Your last line is wrong, because you only proved $f(a_n) \to f(a)$ for a very special type sequence... $\endgroup$ – N. S. Dec 17 '16 at 22:05
  • $\begingroup$ Thanks! Fixed! But the idea keeps the same! $\endgroup$ – Arnaldo Dec 17 '16 at 22:24
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$f$ is continuous at $0$ if and only if

$$\forall \epsilon>0 \exists \delta>0 : |x|<\delta \implies |f(x)-f(0)|<\epsilon.$$

Since $f(0)=f(0+0)=f(0)+f(0)=2f(0)$ we have $f(0)=0.$ Thus

$f$ is continuous at $0$ if and only if

$$\forall \epsilon>0 \exists \delta>0 : |x|<\delta \implies |f(x)|<\epsilon.$$

Now, $f$ is continuous at $x_0$ if and only if

$$\forall \epsilon>0 \exists \delta>0 : |h|<\delta \implies |f(x_0+h)-f(x_0)|<\epsilon.$$ This obviously holds since $$f(x_0+h)-f(x_0)=f(h)$$ and $f$ is continuous at $0.$

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