Question

Suppose f satisfies the relation f(x+y)=f(x)+f(y) and f continuous at 0. Show that f is continuous at all points.

Answer 1

\[f(x+y)=f(x)+f(y)\\
f(x+y)-f(x)=f(y)\\
\frac{f(x+y)-f(x)}{y}=\frac{f(y)}{y}, y \neq 0\\
\lim_{y \to 0}\frac{f(x+y)-f(x)}{y}=\lim_{y \to 0} \frac{f(y)}{y}\\
f'(x)=\lim_{y \to 0} \frac{f(y)}{y}\]
If RHS exists, then f is differentiable and therefore continuous.

Answer 2

do we need to use epsilon delta?

Answer 3

Where would you use it?

Answer 4

and f(x)=f(x+0)=f(x)+f(0),so we get f(0)=0.

Answer 5

Then use L'Hospital's rule.

Answer 6

we can try this
\[\lim_{x\to a}f(x)=\lim_{h\to 0}f(a+h)\]

Answer 7

First we show that\[f(0) = 0\]This is true because:\[f(0)=f(0+0)=f(0)+f(0)=2f(0)\Longrightarrow f(0)=0\]Now we how that \[f(-x)=-f(x)\]This is because:\[0=f(0)=f(x-x)=f(x)+f(-x)\Longrightarrow -f(x)=f(-x)\]Now, to show that f is continuous at a point c, we need to show that:\[\lim_{x\rightarrow c}f(x)=f(c)\]We know that f is continuous at 0, so we have:\[\lim_{x\rightarrow 0}f(x)=f(0)=0\]Let c be any real number. Then:\[\lim_{x\rightarrow 0}f(x)=0\Longrightarrow \lim_{x\rightarrow c}f(x-c)=0\]Using the properties we proved, we obtain:\[\lim_{x\rightarrow c}f(x-c)=0\Longrightarrow \lim_{x\rightarrow c}\left(f(x)-f(c)\right)=0\]\[\Longrightarrow \lim_{x\rightarrow c}f(x)-\lim_{x\rightarrow c}f(c)=0\]\[\Longrightarrow \lim_{x\rightarrow c}f(x)-f(c)=0\Longrightarrow \lim_{x\rightarrow c}=f(c)\]Hence f is continuous at c, and because c is arbitrary, its continuous over the reals.

Answer 8

This can be written in a more rigorous manner using epsilon-delta, however the main idea is there.

Answer 9

I had actually worked out this problem a few days ago. its number 12 in section 5.2 of Introduction to Real Analysis by Bartle.

Answer 10

oo yeah..i understand..thanks a lot...

Answer 11

how we get \[ \lim_{x \rightarrow c} f(c) = f(c)\] ?

Answer 12

The statement:\[f(c)\]is just a constant. It doesnt depend on x. Its like asking:\[\lim_{x\rightarrow 0} 5\]you would just get 5 since the function doesnt depend on x.

Answer 13

ok2...

Answer 14

wow! what i get for going to eat. nice work!
hi joe

Answer 15

here is my proof. see if it works
\[\lim_{x\to a}f(x)=\lim_{h\to 0}f(a+h)\]
\[=\lim_{h\to 0}f(a)+f(h)=f(a)+\lim_{h\to 0}f(h)=f(a)+f(0)=f(a)\]