$f(x)$ is integrable in every interval in $\mathbb{R}$ and $\forall \ x \in \mathbb{R}$, $f(x)=\int_{0}^{x}{f(t)}dt$
I need to prove that $f(x)=0 \ \forall \ x$
Can someone give a clue? Thanks!
Asked
Active
Viewed 99 times
1
-
1Welcime to MSE, please use Mathjax to type your question. And also please add at least what have you tried so that others can help you. – Nizar Mar 16 '16 at 09:36
-
See also this question. – Martin Sleziak Jan 07 '17 at 19:07
2 Answers
2
Hint:
What's the derivative of $f$?
5xum
- 123,496
- 6
- 128
- 204
-
-
-
-
@david Well, what's the family of functions that satisfies the property "the derivative of the function is the function itself"? – 5xum Mar 16 '16 at 10:00
-
-
@david Are you sure you know about differential equations? What about $e^x$? and $2e^x$? – 5xum Mar 16 '16 at 10:10
-
-
@david Correct. Now, you know that $f(x)=C\cdot e^x$ for some $C\in\mathbb R$. You need to determine $C$. To do that, look at the value of $f(0)$. – 5xum Mar 16 '16 at 10:34
-
-
@david If this answer is what you needed, you can also upvote and accept it. This is how the site works. – 5xum Mar 16 '16 at 10:59
0
Actually you have to notice this: the function $f$ is only assumed to be integrable, so a priori you can't say anything about the derivative of $\int_{0}^x f(t)dt$, unless you notice that the fundamental theorem of calculus says you that $f(x) = \int_{0}^x f(t)dt$ is continous at every point, now you can apply further the fundamental theorem of calculus to calculate that $f' = f$ and the claim easily follows as discussed in comments
jJjjJ
- 429
- 4
- 11