Let \(P_2\) be the real polynomials of degree at most two and define \[ J(f)=\int_0^1 f(x)^2\,dx. \] Let \[ Q=\{f\in P_2:f(1)=1\}. \] Show that \(J\) attains a minimum on \(Q\) and determine the minimizing polynomial.
Solution
Equip \(P_2\) with the inner product \[ \langle p,q\rangle \coloneqq \int_0^1p(x)q(x)\,dx. \] Then \(J(f)=\langle f,f\rangle\).
1. For\[ k(x)\coloneqq3-24x+30x^2, \]one has\[ \langle q,k\rangle=q(1) \]for every \(q\in P_2\).
It is enough to check the basis \(1,x,x^2\). Directly, \[ \int_0^1k(x)\,dx = 3-12+10 = 1, \] \[ \int_0^1xk(x)\,dx = \frac32-8+\frac{15}{2} = 1, \] and \[ \int_0^1x^2k(x)\,dx = 1-6+6 = 1. \] Thus the two linear functionals \[ q\longmapsto\langle q,k\rangle \qquad\text{and}\qquad q\longmapsto q(1) \] agree on a basis of \(P_2\), hence agree everywhere.
2. The squared norm of \(k\) is\[ \langle k,k\rangle=9. \]
Apply step 1 with \(q=k\). Then \[ \langle k,k\rangle=k(1)=3-24+30=9. \]
3. Every \(f\in Q\) satisfies\[ J(f)\ge\frac19. \]
If \(f\in Q\), then \(f(1)=1\). By step 1, \[ 1 = \langle f,k\rangle. \] Cauchy–Schwarz and step 2 give \[ 1 \le \langle f,f\rangle^{1/2}\langle k,k\rangle^{1/2} = 3J(f)^{1/2}. \] Squaring yields \(J(f)\ge1/9\).
4. Equality in step 3 occurs for exactly one polynomial, namely\[ f_0(x) \coloneqq \frac{k(x)}9 = \boxed{\frac{10x^2-8x+1}{3}}. \]
Equality in Cauchy–Schwarz holds exactly when \(f\) and \(k\) are linearly dependent. Thus an equality case in \(Q\) must have \(f=ck\). The constraint \(f(1)=1\) and \(k(1)=9\) force \(c=1/9\), giving the displayed polynomial. Conversely, \[ f_0(1)=\frac{10-8+1}{3}=1, \] so \(f_0\in Q\), and step 2 gives \[ J(f_0) = \frac1{81}\langle k,k\rangle = \frac19. \]
5. Therefore \(J\) attains its minimum on \(Q\), with\[ \boxed{ \min_{f\in Q}J(f)=\frac19, \qquad f_{\min}(x)=\frac{10x^2-8x+1}{3}. } \]
Step 3 gives the universal lower bound, and step 4 gives the unique element of \(Q\) attaining it.
6. Q.E.D.
Step 5 gives both the existence and the location of the minimum.