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 September 28th, 2011, 09:40 PM #1 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 On Continuous, Integrable Functions on [0,1] Let $0 \leq a \leq 1$ be fixed. Determine all non-negative continuous functions $f:[0,1]\to \mathbb{R}$ which satisfy the following three conditions: 1.) $\int_0^1 f(x)\,dx= 1$, 2.) $\int_0^1 x f(x)\,dx= a$, 3.) $\int_0^1 x^2 f(x)\,dx= a^2$. First of all, I have no idea how to begin this problem. We were told that we can use general facts from Calculus without rigor to prove this, but even with that I cannot seen to find a place to begin.
 September 29th, 2011, 09:26 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: On Continuous, Integrable Functions on [0,1] You would agree, I hope, that if $f$ is nonnegative then the function $(x-a)^2f(x)$ is nonnegative everywhere. What happens when you evaluate its integral over [0,1], and what does this tell you about $f?$
 September 29th, 2011, 03:15 PM #3 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: On Continuous, Integrable Functions on [0,1] I believe it is no-negative, but I am not really sure I follow you on the rest. The trouble I seem to be having is integrating an undefined function f(x) over a definite integral.
 September 30th, 2011, 02:43 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: On Continuous, Integrable Functions on [0,1] If we expand $(x-a)^2f(x),$ we get $x^2f(x)-2axf(x)+a^2f(x).$ You know what you get when you integrate $f(x),\ xf(x)$ and $x^2f(x)$ over the interval [0,1], so you should be able to work out what the integral of the above function is over the same interval without knowing anything more explicit about f.
 September 30th, 2011, 08:08 AM #5 Newbie   Joined: Dec 2008 From: Copenhagen, Denmark Posts: 29 Thanks: 0 Re: On Continuous, Integrable Functions on [0,1] Multiply 1) with a squared and 2) with a. Then set them equal and you reach the point mattpi is making.
 September 30th, 2011, 08:25 AM #6 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: On Continuous, Integrable Functions on [0,1] Oh, I see what you are saying. Are we looking for a contradiction?
 September 30th, 2011, 09:22 AM #7 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: On Continuous, Integrable Functions on [0,1] What do we know about a continuous function if it integrates to zero over an interval, and is non-negative in that same interval?
 September 30th, 2011, 09:27 AM #8 Newbie   Joined: Dec 2008 From: Copenhagen, Denmark Posts: 29 Thanks: 0 Re: On Continuous, Integrable Functions on [0,1] I wouldn't call it a contradiction, but an implication. Let's see, we have $\int_0^1 a^2 f(x) dx= \int_0^1 ax f(x) dx = \int_0^1 x^2 f(x) dx,$ thus $\int_0^1 (a^2 - ax) f(x) dx= 0,$ and $\int_0^1 (x^2 - ax) f(x) dx= 0,$ which means that $\int_0^1 (x^2 - 2ax + a^2) f(x) dx= 0 \quad \Leftrightarrow \quad \int_0^1 (x-a)^2 f(x) dx = 0.$ Knowing that $(x-a)^2$ is non-negative for all $x \in [0,1]$, this last integral equation implies something about $f$. What does it imply?
 September 30th, 2011, 09:37 AM #9 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: On Continuous, Integrable Functions on [0,1] It must be that f is the zero function. But that contradicts 1) So no such functions exist.

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