Let \(S\) be the set of real \(C^1\) functions \(f\) on \([0,1]\) such that \[ f(0)=0, \qquad \int_0^1 f'(x)^2\,dx\le1. \] Define \[ J(f)=\int_0^1 f(x)\,dx. \] Show that \(J\) is bounded on \(S\) and compute its supremum. Determine whether the maximum is attained, and if so find a maximizing function.
Solution
1. For every \(f\in S\),\[ J(f)=\int_0^1(1-x)f'(x)\,dx. \]
Integration by parts gives \[ \begin{aligned} \int_0^1 f(x)\,dx &= \left[(x-1)f(x)\right]_0^1 -\int_0^1(x-1)f'(x)\,dx\\ &= \int_0^1(1-x)f'(x)\,dx, \end{aligned} \] because \(f(0)=0\) and the factor \(x-1\) vanishes at \(x=1\).
2. The functional \(J\) is bounded on \(S\), and\[ \abs{J(f)}\leq\frac{1}{\sqrt3} \]for every \(f\in S\).
By step 1 and the Cauchy–Schwarz inequality, \[ \begin{aligned} \abs{J(f)} &\leq \left(\int_0^1(1-x)^2\,dx\right)^{1/2} \left(\int_0^1f'(x)^2\,dx\right)^{1/2}\\ &\leq \left(\frac13\right)^{1/2} = \frac{1}{\sqrt3}. \end{aligned} \] The second factor is at most \(1\) by the defining condition for \(S\).
3. Define\[ f_0(x)=\sqrt3\left(x-\frac{x^2}{2}\right). \]Then \(f_0\in S\) and\[ J(f_0)=\frac{1}{\sqrt3}. \]
One has \(f_0(0)=0\) and \[ f_0'(x)=\sqrt3(1-x). \] Therefore \[ \int_0^1 f_0'(x)^2\,dx = 3\int_0^1(1-x)^2\,dx = 1, \] so \(f_0\in S\). By step 1, \[ J(f_0) = \sqrt3\int_0^1(1-x)^2\,dx = \frac{\sqrt3}{3} = \frac{1}{\sqrt3}. \]
4. The supremum of \(J\) on \(S\) is\[ \boxed{\frac{1}{\sqrt3}}, \]and this value is attained at\[ \boxed{f_0(x)=\sqrt3\left(x-\frac{x^2}{2}\right)}. \]
Step 2 gives \[ J(f)\leq\frac{1}{\sqrt3} \] for every \(f\in S\), while step 3 exhibits \(f_0\in S\) with equality. Hence the upper bound is the supremum and is attained by \(f_0\).
5. The maximizing function is unique.
2. Equality in Cauchy–Schwarz gives\[ f'(x)=c(1-x) \]for some constant \(c\geq0\), and equality in the norm bound gives\[ 1 = \int_0^1f'(x)^2\,dx = \frac{c^2}{3}. \]Thus \(c=\sqrt3\). Since \(f(0)=0\), integration yields\[ f(x)=\sqrt3\left(x-\frac{x^2}{2}\right)=f_0(x). \]
6. Q.E.D.
Steps 2 and 4 prove boundedness, identify the supremum, and exhibit a maximizing function; step 5 gives uniqueness.