Let, $f:\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ be a function satisfying $f(x+y)=f(x)f(y)$. Given that, $f$ is continuous a $x=0$. Prove that, f is continuous on $\mathbb{R}$.
Asked
Active
Viewed 103 times
-1
-
3did you try anything at all? – Ittay Weiss Dec 19 '13 at 18:15
-
1Notice that you can translate a point $a$ to the origin by using $f(x+a)$, but because of the given property the function looks again at the origin since $f(x+a)=f(a)f(x)$. Notice that $f(a)$ is a constant and $f(x)$ is at the origin. – OR. Dec 19 '13 at 18:17
-
Don't worry about bad professors making you waste your time asking if you have tried it. Shame on them. – OR. Dec 19 '13 at 18:18
-
After seeing how this problem was solved, pay attention on what features of the property $f(x+y)=f(x)f(y)$ made it possible for the technique used in the solution to work. You can try then similar-looking problems. For example, replace in your problem the given equation by $f(x+y)=f(x)+f(y)$. – OR. Dec 19 '13 at 18:24
-
2@ABC We can not move forward, understand mathematics and improve our skills if we don't try: it isn't waste of time and all who do mathematics know this. – Dec 19 '13 at 18:27
-
It is a waste of time asking it here. You can imagine that after every post saying "did you try" there is another saying "yes, but I didn't know what to do". After all, the OP already went out of his/her way to find a solution. Part of the learning process is also seeing solutions. – OR. Dec 19 '13 at 18:31
-
The only reason people ask "did you try it" is because they feel they may be solving someone's homework, and they have the false believe that that is a bad thing, that homework should be individual. – OR. Dec 19 '13 at 18:33
-
2@ABC Ok but at same time there's another who think to post a question without trying in it and by seeing this question he know that he must do some effort. – Dec 19 '13 at 18:36
-
Sure. "did you eat enough vegetables today?". It is equally important. I believe more that the intention is encouraging effort, when a hint is given. Those that ask for showing effort have other intentions (bossing, trying to prevent people from getting homework answers). I don't hold any of those in high esteem. – OR. Dec 19 '13 at 18:48
-
@ABC I saw your profile where you have also written on this subject, ok it is a point of view that I respect and even I can understand it. – Dec 19 '13 at 18:58
-
1I'd just like to point out that it is absolutely false that the only reason to ask if someone has tried anything is the belief that solving someone's homework is a bad thing. The answer to that question can be very useful in determining how someone's thinking about a problem, why that thinking has been unfruitful or led then astray, and what exactly needs to be explained in order for them to understand. – Jim Dec 19 '13 at 19:33
-
1Assuming that every time someone asks that question they're just being snobby about making someone do their homework is every bit as snobby and unhelpful as what you've assumed the other person is doing. – Jim Dec 19 '13 at 19:35
-
can you tell me what is the geometry of this type of function $f(x+y)=f(x)f(y)$ – Topology Dec 19 '13 at 19:36
-
@Mathematics What do you mean by the geometry of this type of function? – Dec 19 '13 at 19:39
-
I mean is there any geometrical behaviour of this type of functions? @ Sami Ben Romdhane – Topology Dec 19 '13 at 19:43
-
We can prove that the functions that satisfy this functional equality are either the zero function or the exponential functions. – Dec 19 '13 at 19:44
-
how can we prove? @ Sami Ben Romdhane – Topology Dec 19 '13 at 19:51
-
@Mathematics Ask this question for the benefit be for everyone. – Dec 19 '13 at 20:06
-
Please make it clear how to do..I am not being able to do this @ Sami Ben Romdhane – Topology Dec 19 '13 at 20:08
