Find $\int_{0}^{1} f(x)^3\:dx$

Question

If $f:[0,1]\to R$ satisfies: $$\int_{0}^{1}f(x)dx=1$$ $$\int_{0}^{1}xf(x)dx=1$$ $$\int_{0}^{1}f(x)^2dx=4$$ Find: $$\int_{0}^{1}f(x)^3dx$$

My try: Let $f(1)=p$

I used Parts: we get

$$p-\int_{0}^{1}xf'(x)dx=1$$

$$p-\int_{0}^{1}(xf'(x)+f(x))xdx=1$$ i.e., $$p-2=\int_{0}^{1}x^2f'(x)dx$$

Also:

$$p^2-2\int_{0}^{1}f(x)f'(x)x\:dx=4$$

i am stuck here?

In general if those three integrals are given, like $\int_0^1 f(x)dx=a,\ \int_0^1 xf(x)dx=b,\ \int_0^1 f^2(x)dx=c$ one can consider the following integral $\int_0^1 (f(x)+mx+n)^2dx$ and set it to $0$ in order to find $m$ and $n$, of course it won't be always easy to find them. — Zacky, Jan 24 '20 at 21:03
@Zacky Arbitrary values of $a,b,c$ don't allow for such $m,n$. See my answer to see why exactly they existed in this specific question. For other values the information given might not be enough to determine the almost-everywhere class of $f$. — OscarRascal, Jan 24 '20 at 21:09

score 5 · Accepted Answer · answered Jan 24 '20 at 20:11

5

Hint :

$$ \int_0^1 \left\{f(x)-6x+2 \right\}^2dx = 0$$

answered Jan 24 '20 at 20:11

bFur4list

2,604

You made my day awesome HINT, but not sure how did you come over with this idea – Ekaveera Gouribhatla Jan 24 '20 at 20:16
@Zacky $f(x)=6x-2$ clearly satisfies all the equations – Ekaveera Gouribhatla Jan 24 '20 at 20:20
@Zacky The integrand is non-negative and the integral is zero. Therefore, the integrand must be zero almost everywhere. It follows that $f^3(x)=(6x-2)^3$ almost everywhere. From there $\int_{0}^{1}f^3(x)dx=\int_{0}^{1}(6x-2)^3dx=10$. – OscarRascal Jan 24 '20 at 20:22
@EkaveeraKumarSharma Yes, but the real point is that a square is non-negative, so we must have $f(x)=6x-2$ almost everywhere in the range. – almagest Jan 24 '20 at 20:22

OscarRascal · Answer 2 · 2020-01-24T21:13:17.300

Let me give you a deductive solution by linear algebra.

Define $$\begin{align}\langle h_1,h_2\rangle&=\int_{0}^{1}h_1(x)h_2(x)dx\\\|h\|^2&=(h,h)\end{align}$$

The information that is given is that $$\begin{align}\langle 1,f\rangle&=1\\\langle x,f\rangle&=1\\\langle f,f\rangle&=4\end{align}$$

Let's orthonormalize the system $1,x$. This is, we keep $f_1(x)=1$ and translate $x$ by a multiple of $f_1$ to get a function $f_2$ such that $\langle f_1,f_2\rangle=0$ and $\|f_2\|=1$.

We need that $0=(1,x+a)=\int_{0}^{1}(x+a)dx=\frac{1}{2}+a$. Therefore, $a=-\frac{1}{2}$.

Now, $\langle x-1/2,x-1/2\rangle=\frac{1}{12}$.

Therefore, $f_1(x)=1,f_2(x)=2\sqrt{3}\left(x-\frac{1}{2}\right)$ is orthonormal.

Note that $\langle f_2,f\rangle=\sqrt{12}\left(\langle x,f\rangle-\frac{1}{2}\langle 1,f\rangle\right)=\sqrt{3}$.

Then $\langle f,f\rangle=4=1^2+(\sqrt{3})^2=|\langle f_1,f\rangle|^2+|\langle f_2,f\rangle|^2$.

So, these functions satisfy the equality in Bessel's inequality.

We can now use the proof of Bessel's inequality

$$\begin{align}0&=\langle f,f\rangle -\langle f_1,f\rangle ^2-\langle f_2,f\rangle ^2\\ &=\langle f,f\rangle -2[\langle f_1,f\rangle ^2+\langle f_2,f\rangle ^2]+[\langle f_1,f\rangle ^2+\langle f_2,f\rangle ^2]\\ &=\|f-\langle f_1,f\rangle f_1-\langle f_2,f\rangle f_2\|^2\end{align}$$ to conclude that $f$ is $\langle f_1,f\rangle f_1-\langle f_2,f\rangle f_2$ almost everywhere and compute from it $\int_{0}^{1}f^3(x)dx$.

Of course, $\|f-\langle f_1,f\rangle f_1-\langle f_2,f\rangle f_2\|^2=0$ is the hint that you were given, since $$\langle f_1,f\rangle f_1+\langle f_2,f\rangle f_2=1+6x-3=6x-2,$$ and now you can see where it comes from.

Find $\int_{0}^{1} f(x)^3\:dx$

2 Answers2