# No non-negative continuous function on $[a,b]$ such that $\int_a^b f(t)dt=1, \int_a^b tf(t)dt=c, \int_a^b t^2f(t)dt=c^2$ for $c\in\mathbb{R}$.

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!

## 2 Answers

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 $$t \in [a, b]$$.

• What is a quick explanation that $c > 0$? – Umberto P. Mar 5 at 19:44
• Actually, we don't need that. Edited! – Tom Chen Mar 5 at 20:10
• 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 $c = k$. – Tom Chen Mar 5 at 20:14
• The Dirac function is not continuous, it isn't even a function. ;-) – Mars Plastic Mar 5 at 21:01
• @MarsPlastic Ha! Yes, I should read the prompt more carefully ;) – Tom Chen Mar 6 at 0:34

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?