# If $f(x+y)=f(x)+f(y) ,\forall\;x,y\in\Bbb{R}$, then if $f$ is continuous at $0$, then it is continuous on $\Bbb{R}.$ [duplicate]

I know that this question has been asked here before but I want to use a different approach. Here is the question.

A function $f:\Bbb{R}\to\Bbb{R}$ is such that \begin{align} f(x+y)=f(x)+f(y) ,\;\;\forall\;x,y\in\Bbb{R}\qquad\qquad\qquad(1)\end{align} I want to show that if $f$ is continuous at $0$, it is continuous on $\Bbb{R}.$

MY WORK

Since $(1)$ holds for all $x\in \Bbb{R},$ we let \begin{align} x=x-y+y\end{align} Then, \begin{align} f(x-y+y)=f(x-y)+f(y)\end{align} \begin{align} f(x-y)=f(x)-f(y)\end{align} Let $x_0\in \Bbb{R}, \;\epsilon>$ and $y=x-x_0,\;\;\forall\,x\in\Bbb{R}.$ Then, \begin{align} f(x-(x-x_0))=f(x)-f(x-x_0)\end{align} \begin{align} f(x_0)=f(x)-f(x-x_0)\end{align} \begin{align} f(y)=f(x_0)-f(x)\end{align}

HINTS BY MY PDF:

Let $x_0\in \Bbb{R}, \;\epsilon>$ and $y=x-x_0,\;\;\forall\,x\in\Bbb{R}.$ Then, show that \begin{align} \left|f(x_0)-f(x)\right|=\left|f(y)-f(0)\right|\end{align} Using this equation and the continuity of $f$ at $0$, establish properly that \begin{align}\left|f(y)-f(0)\right|<\epsilon,\end{align} in some neighbourhood of $0$.

My problem is how to put this hint together to complete the proof. Please, I need assistance, thanks!

## marked as duplicate by dxiv, Chase Ryan Taylor, Jyrki Lahtonen, user91500, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 7 '18 at 11:07

• we let x=x−y+y Guess you mean $\,x \mapsto x-y\,$ there, instead. – dxiv Sep 7 '18 at 5:26
• Prove that $f(y)=f(y)-f(0)$, and you're basically there. Then you have to prove the hint itself too, of course. – Arthur Sep 7 '18 at 5:26

We want to show that

$$\forall \epsilon>0, \exists r>0:|x-y| <r \implies |f(x) - f(y)| < \epsilon$$

But $f(x)-f(y)=f(x-y)$ because $f(y)+f(x-y)=f(y+(x-y))=f(x)$ as you have noticed.

Now, take $u=x-y$. By continuity at $0$, we can write:

$$\forall \epsilon>0, \exists r>0:|u-0| <r \implies |f(u) - f(0)| < \epsilon$$

It's easy to see that $f(0)=0$, because $f(0)=f(0+0)=f(0)+f(0)$. Hence

$$\forall \epsilon>0, \exists r>0:|(x-y)-0| <r \implies |f(x-y) - 0| < \epsilon$$ $$\forall \epsilon>0, \exists r>0:|x-y| <r \implies |f(x)-f(y)| < \epsilon$$ Hence, $f$ is continuous at any $y \in \mathbb{R}$.

• Thanks a lot, I am grateful! – Micheal Sep 7 '18 at 7:26
• @Micheal You're welcome. – stressed out Sep 7 '18 at 7:27

One way without that hint is limiting function as $x\to0$ then $$\lim_{x\to0}f(x+a)=\lim_{x\to0}f(x)+\lim_{x\to0}f(a)=0+f(a)=f(a)$$ because $f$ is continuous at $x=0$. Now let $x+a=t$ then $$\lim_{t\to a}f(t)=f(a)$$

• Thanks, I am grateful for this! – Micheal Sep 7 '18 at 7:27