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Find a real function $w(t)\in L_2[0,1]$ such that:

  • $w(t)\geq 0 \quad \forall t\in [0,1];$

  • $\displaystyle\int_0^{s}w(t)dt\leq s \quad \forall s\in [0,1];$

  • $\displaystyle \int_0^1 w(t)dt\leq 2;$

  • $\displaystyle \int_0^1 tw(t)dt= 1.$

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1  
Thanks to @Ragib Zaman for rescuing me from a terrible terrible conceptual error. –  Christopher A. Wong Oct 4 '12 at 6:37
    
Thank you for all your comments and helping. I would like to ask more. Let $w(t)\in L_2[0,1]$ be a function such that $$w(t)\geq 0 \quad\forall t\in [0,1],$$ $$\int_o^s w(t)dt\leq s \quad \forall s\in [0,1].$$ Find the minimum of the value $\displaystyle\int_0^1 tw(t)dt$. –  blindman Oct 4 '12 at 7:38
    
Are those the only conditions? Why can't we just take $w \equiv 0$? –  Christopher A. Wong Oct 4 '12 at 7:45
    
@Christopher A. Wong: I am sorry. "minimum" should be replaced by "maximize". –  blindman Oct 4 '12 at 7:49
    
Take a look at the solutions offered by people below, and see if you can determine the answer to that. Hint: You know that you can't attain exactly 1. –  Christopher A. Wong Oct 4 '12 at 7:59

3 Answers 3

up vote 2 down vote accepted

$w(t)$ cannot exist - Since $w(t)$ must be nonzero on some set of positive measure, then $$ 1 = \int_0^1 t w(t) \, dt < \int_0^1 w(t) \, dt \le 1$$ which is a contradiction.

P.S. This was edited from a previously wrong solution in which I (incorrectly) differentiated an inequality!

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Nice: +1 (of course) –  Fabian Oct 4 '12 at 6:13
    
Actually right now I'm trying to make sure I'm not wrong here; I can't quite seem to remember how readily we can use the fundamental theorem of calculus in the context of measurable functions. –  Christopher A. Wong Oct 4 '12 at 6:17
    
Oh, right, the Lebesgue differentiation theorem does hold for any $L^p_{\mathrm{loc}}$. –  Christopher A. Wong Oct 4 '12 at 6:19
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Differentiating inequalities generally is not valid. –  Ragib Zaman Oct 4 '12 at 6:23
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It's best to think of it visually. Functions which oscillate may have similar values to a more regular function but have greatly different derivatives. E.g., $\sin x > -2$ –  Ragib Zaman Oct 4 '12 at 6:33

Let $s=1$ in (2), we have $\int_0^1 w(t)dt \leq 1$. But from (4) we have $1=\int_0^1 t w(t)dt \leq \int_0^1 w(t)dt \leq 1$. So it is true that $\int_0^1 t w(t)dt = \int_0^1 w(t)dt = 1$. The only possibility for $w(t)$ is $$w(t)=\delta(t) \text{ which is the Dirac delta function}$$

If you think $w(t)$ as a probability density function for a random variable $X$ which has bounded support on $[0,1]$, then $X\in [0,1]$ and $EX=1$. $X$ has to be degenerate at $x=1$.

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Note that $\int_0^1 t w(t)$ should be equal to 1 and not smaller than 1. –  Fabian Oct 4 '12 at 6:05
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Why does it matter that $w(1) = 1$? None of the conditions in the above can distinguish values on null sets. –  Christopher A. Wong Oct 4 '12 at 6:07
    
@Fabian Yes, fix that. Thanks. –  Patrick Li Oct 4 '12 at 6:08
    
I'm also confused how your example gives $\int_0^1tw(t)dt=1$. –  Fabian Oct 4 '12 at 6:11
    
It should be the Dirac delta function here. –  Patrick Li Oct 4 '12 at 6:12

No such functions exist. To see this, let $f(s) = \displaystyle\int^s_0 w(t) dt.$ The first condition reads $f(s)\leq s$ which in particular gives $f(1)\leq 1.$ Note also that this makes the second condition redundant.

Integrate the third condition by parts to get to $$ f(1) - \int^1_0 f(t) dt = 1.$$

Since $f(1)\leq 1$ we have $$\int^1_0 f(t) dt = f(1) - 1 \leq 0.$$

However, $f(t)$ must be non-negative as $w(t)$ is, thus we could only possibly have $$\int^1_0 f(t) dt =0 $$ and this only in the case where $f(t)=0$ almost everywhere, which would imply that $w(t)=0$ almost everywhere, contradicting the third condition.

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I think there might be a slight mistake in your answer. Note that $f(x)$ is the anti-derivative of $w(x)$ in the beginning and later on becomes equivalent to $w(x)$. –  Fabian Oct 4 '12 at 5:51
    
@Fabian I can't see where that occurs, can you please elaborate? –  Ragib Zaman Oct 4 '12 at 6:20
    
@Ragib Zaman: Thank you for your solution. I would like to ask you more in the above comment. –  blindman Oct 4 '12 at 7:41
    
@RagibZaman: My question in more details. How do you get from the third condition to the condition $1=f(1) - \int_0^1f(t) dt= \int_0^1 w(t)dt - \int_0^1 \int_0^t w(s) ds dt$? –  Fabian Oct 4 '12 at 12:03
    
@Fabian Let $u=t$ and $dv = w(t) dt$ with the aiming of applying the integration by parts formula $ \int^b_a u dv = uv\mid^b_a - \int^b_a v du.$ Since $u=t$, we have $du=dt.$ By definition of $f$ we also have $v= f(t).$ Then the formula gives $$ \int^1_0 t w(t) dt = tf(t)\mid^1_0 - \int^1_0 f(t) dt = f(1) - \int^1_0 f(t) dt.$$ –  Ragib Zaman Oct 4 '12 at 12:22

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