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Show that there exists no non-negative continuous function $f$ defined on the interval $[a,b]$ such that it satisfies the following conditions: $$\int_a^b f(t)dt=1 \quad \int_a^b tf(t)dt = c \quad \int_a^b t^2f(t)dt = c^2,$$ for some $c\in\mathbb{R}$.

I was given a hint that I need to use Cauchy-Schwarz inequality, and I am familiar with Cauchy-Schwarz, I just do not see how to apply it to this problem. Cauchy-Schwarz inequality states that if $u$ and $v$ are elements of an inner product space then, $\| \langle u,v \rangle \| \leq \| u \| \| v \|$.

So I guess here I have the inner product space of $C([a,b])$, continuous functions on $[a,b]$, and I have that $$\langle f, 1 \rangle = 1 \quad \langle f,t\rangle = c \quad \langle f,t^2 \rangle = c^2$$ I don't know how to piece it together though. Any help would be appreciated!

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I'm assuming that $a \neq b$. Then, \begin{align*} |c| &= \left|\int_a^b tf(t) dt\right| \\ &= \left|\int_a^b t \sqrt{f(t)} \sqrt{f(t)} dt\right| \\ &\le \left(\int_a^b t^2 f(t) dt\right)^{1/2}\left(\int_a^b f(t) dt\right)^{1/2} \\ &= (c^2)^{1/2}\cdot 1^{1/2} \\ &= |c| \end{align*} Of course, the equality must hold, otherwise we have a contradiction. The equality condition in Cauchy-Schwarz is when $t \sqrt{f(t)} \propto \sqrt{f(t)}$, or $t \propto 1$, which is impossible for $t \in [a, b]$.

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  • $\begingroup$ What is a quick explanation that $c > 0$? $\endgroup$ – Umberto P. Mar 5 at 19:44
  • $\begingroup$ Actually, we don't need that. Edited! $\endgroup$ – Tom Chen Mar 5 at 20:10
  • $\begingroup$ Also, why I assume at $a\neq b$ is because if $a = b \overset{\text{def}}{=} k$, then we could take $f(t) = \delta_k(t)$, where $\delta_x(t)$ is the delta Dirac function. Then, we could take $c = k$. $\endgroup$ – Tom Chen Mar 5 at 20:14
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    $\begingroup$ The Dirac function is not continuous, it isn't even a function. ;-) $\endgroup$ – Mars Plastic Mar 5 at 21:01
  • $\begingroup$ @MarsPlastic Ha! Yes, I should read the prompt more carefully ;) $\endgroup$ – Tom Chen Mar 6 at 0:34
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Hint: Check that $$ \langle u,v\rangle_f:=\int_a^b u(t)v(t)f(t)dt \quad \text{for all $u,v\in C([a,b])$}$$ defines an inner product. Then use that by assumption $$ \langle 1,t\rangle_f ^2 = \langle 1,1\rangle_f \langle t,t\rangle_f.$$ What do you know about the case in which equality holds in the Cauchy-Schwarz inequality?

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