Vtyjkyuk

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

A function $f:BbbRtoBbbR$ is such that
beginalign f(x+y)=f(x)+f(y) ,;;forall;x,yinBbbRqquadqquadqquad(1)endalign
I want to show that if $f$ is continuous at $0$, it is continuous on $BbbR.$

MY WORK

Since $(1)$ holds for all $xin BbbR,$ we let beginalign x=x-y+yendalign
Then,
beginalign f(x-y+y)=f(x-y)+f(y)endalign
beginalign f(x-y)=f(x)-f(y)endalign
Let $x_0in BbbR, ;epsilon>$ and $y=x-x_0,;;forall,xinBbbR.$ Then,
beginalign f(x-(x-x_0))=f(x)-f(x-x_0)endalign
beginalign f(x_0)=f(x)-f(x-x_0)endalign
beginalign f(y)=f(x_0)-f(x)endalign

HINTS BY MY PDF:

Let $x_0in BbbR, ;epsilon>$ and $y=x-x_0,;;forall,xinBbbR.$ Then, show that beginalign left|f(x_0)-f(x)right|=left|f(y)-f(0)right|endalign
Using this equation and the continuity of $f$ at $0$, establish properly that
beginalignleft|f(y)-f(0)right|<epsilon,endalign
in some neighbourhood of $0$.

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

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 $y in mathbbR$.

One way without that hint is limiting function as $xto0$ then
$$lim_xto0f(x+a)=lim_xto0f(x)+lim_xto0f(a)=0+f(a)=f(a)$$
because $f$ is continuous at $x=0$. Now let $x+a=t$ then
$$lim_tto af(t)=f(a)$$