# Proof of bound on $\int t\,f(t)\ dt$ given well-behaved $f$

I got the following question by mail from someone I don't know from Adam. (Quoted in part.)

if $f(t)$ continuously diff. on $[0,1]$ and

a) $\int_0^1f(t)\ dt=0$

b) $m\le f\,'\le M$ on $[0,1]$

Prove

$\frac m{12}\le\int_0^1t\cdot f(t)\ dt\le\frac M{12}$

I suspect it might be an error

I assumed immediately that it's an error, but my first two thoughts as counterexamples were $f(t)=\frac12-t$ and $f(t)=\sin(2\pi t)$, both of which satisfy the result. Anyone with a proof or counterexample?

• Basically the claim is $\frac{m}{12} \leq E(T) \leq \frac{M}{12}$. – PrimeNumber Feb 6 '11 at 23:22
• Tag well-behavedness? (I'm unable create a new tag, but that seems one worth having.) – msh210 Feb 7 '11 at 20:55

## 2 Answers

Edit: Incorporated simplification and shortening of the argument suggested by Didier. Thanks!

By a) we know that $\frac{1}{2} \int_{0}^{1} f(x)\,dx = 0$. Therefore \begin{align*} \int_{0}^{1} xf(x)\,dx & = \int_{0}^{1} \left(x - \frac{1}{2}\right)f(x)\,dx \\ & = \left. \frac{1}{2}(x^{2} - x) f(x)\right\vert_{0}^{1} - \int_{0}^{1} \frac{1}{2}(x^{2} - x)f'(x)\,dx \\ & = \frac{1}{2} \int_{0}^{1} (x - x^{2})f'(x)\,dx \end{align*} using integration by parts.

Now note that $x - x^{2} \geq 0$ on $[0,1]$ and by b) we have $m \leq f'(x) \leq M$, hence $$C m \leq \int_{0}^{1} x f(x)\,dx \leq C M$$ with $$C = \frac{1}{2} \int_{0}^{1} (x - x^{2})\,dx = \left.\frac{1}{2}\left(\frac{1}{2}x^{2} - \frac{1}{3}x^{3}\right)\right\vert_{0}^{1} = \frac{1}{12}$$ as we wanted.

Let me add that these estimates arise in the Euler summation method and are often used in proofs of the Stirling formula.

• The detour by $F$ to go from the integral of $xf(x)$ to the integral of $\frac12(x-x^2)f'(x)$ may be omitted once one notes that the former is also the integral of $(x-\frac12)f(x)$, to which one applies the usual integration by parts formula. – Did Feb 7 '11 at 8:14
• Thank you! I knew there was a connection with Euler summation / precise remainder terms for Taylor series, but I couldn't recall the details. – Steven Stadnicki Feb 7 '11 at 8:15
• @Didier: I included it, thanks! – t.b. Feb 7 '11 at 8:27
• You are welcome. – Did Feb 7 '11 at 9:12
• @Didier, beautiful; thanks! – msh210 Feb 7 '11 at 20:54

I presume that the answer comes from integrating $t\cdot f(t)$ by parts; $\int t\cdot f(t)dt = (t\cdot \int_0^t f(x) dx)|_0^1 - \int_0^1(\int_0^t f(x)dx)dt$, and since the first term evaluates to $0$ (the value at $0$ is $0$ because of the factor of $t$, and the value at $1$ is $0$ because $\int_0^1 f(x) dx = 0$) it simplifies to $-\int_0^1(\int_0^t f(x)dx)dt$ ; now, this is $0$ at both ends of the $t$ interval, and so there should be relatively straightforward ways of bounding it.