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This question already has an answer here:

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!

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marked as duplicate by dxiv, Chase Ryan Taylor, Jyrki Lahtonen, user91500, José Carlos Santos real-analysis Sep 7 '18 at 11:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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    $\begingroup$ Thanks a lot, I am grateful! $\endgroup$ – Micheal Sep 7 '18 at 7:26
  • $\begingroup$ @Micheal You're welcome. $\endgroup$ – stressed out Sep 7 '18 at 7:27
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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 $$\lim_{t\to a}f(t)=f(a)$$

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

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