Find maximum value of integrand

Question

For $f$ be continuous functions on $[0,1]$ and satisfies

(i) $\int_{0}^{1} x^2f(x) dx=0$

(ii) $max_{[0,1]}|f(x)|=6$

Find maximum value of $\phi(f)=\int_{0}^{1}x^3f(x)dx$

Help me. Please! Thanks everyone !!

Answer 1

Let $V$ be the real vector spaces of continuous functions on $[0,1]$. Equip $V$ with the inner product $\langle\_,\_\rangle$ defined by $$\langle a,b\rangle := \int_0^1\,a(x)\,b(x)\,\text{d}x$$ for all $a,b\in V$. Moreover, for each integer $n\geq 0$, write $p_n$ for the polynomial function $p_n(x)=x^n$ for all $x\in [0,1]$. Finally, for each function $a \in V$, the orthogonal complement $a^\perp$ is given by $$a^\perp:=\big\{b \in V\,\big|\,\langle a,b\rangle=0\big\}\,.$$

We first look for a function $\psi \in p_2^\perp$ such that $\langle\psi,\psi\rangle=1$ with the property that $\left\langle\psi,p_3\right\rangle$ is maximized. Observe that $q:=p_3-\frac{5}{6} p_2$ lies within $p_2^\perp$. We can instead look for a function $\psi\in p_2^\perp$ with the maximum possible value $\left\langle\psi,q\right\rangle$. By the Cauchy-Schwarz Inequality, $\psi$ must satisfy $\psi=\lambda\,q$ for some $\lambda \in \mathbb{R}_{>0}$. (This can be seen as the intersection of the subspace of $V$ spanned by $q$ with the unit, infinite-dimensional, hypersphere centered at $0$ on $p_2^\perp$.)

Fortunately, to deal with the original question, we do not have to find $\lambda$ explicitly. The required function $\phi$ must be of the form $k\,\psi$ for some $k \in \mathbb{R}_{>0}$, so $\phi=\mu\,q$, where $\mu:=k\lambda\in\mathbb{R}_{>0}$. As the minimum of $q$ is $-\frac{125}{1458}$, whilst the maximum of $q$ equals $\frac16$, we conclude that $\mu=36$, and so $$\phi(x)=36x^3-30x^2$$ for each $x\in[0,1]$.

P.S.: It is possible to shorten this answer considerably by dealing with the integral $$\int_0^1\,\left(x^3-\frac56 x^2\right)\,f(x)\,\text{d}x$$ and apply the Cauchy-Schwarz Inequality directly. However, I am trying to provide a geometric enlightenment.