Let \(f\) and \(f'\) be continuous on \([0,\infty)\), and suppose \[ f(x)=0\qquad(x\ge10^{10}). \] Show that \[ \int_0^\infty f(x)^2\,dx \le 2\sqrt{\int_0^\infty x^2f(x)^2\,dx} \sqrt{\int_0^\infty f'(x)^2\,dx}. \]
Solution
1. One has\[ \left(xf(x)^2\right)' = f(x)^2+2x f(x)f'(x). \]
Apply the product rule and the chain rule: \[ \frac d{dx}\left(xf^2\right) = f^2+x(2ff'). \]
2. One has\[ \int_0^\infty f(x)^2\,dx = -2\int_0^\infty x f(x)f'(x)\,dx. \]
Let \[ R=10^{10}. \] The hypothesis gives \[ f(x)=0 \qquad x\geq R. \] Integrating step 1 over \([0,R]\) gives \[ \left[xf(x)^2\right]_0^R = \int_0^R f(x)^2\,dx +2\int_0^R x f(x)f'(x)\,dx. \] At \(x=R\), \(f(R)=0\), while at \(x=0\) the factor \(x\) makes the boundary term zero. Therefore \[ 0 = \int_0^R f(x)^2\,dx +2\int_0^R x f(x)f'(x)\,dx. \] Since \(f\) is identically zero for \(x\geq R\), one also has \(f'(x)=0\) for \(x>R\). Extending both integrals to \([0,\infty)\) therefore does not change them. Rearranging gives the displayed identity.
3. One has\[ \abs{ \int_0^\infty x f(x)f'(x)\,dx } \leq \sqrt{ \int_0^\infty x^2f(x)^2\,dx } \sqrt{ \int_0^\infty f'(x)^2\,dx }. \]
Apply the Cauchy–Schwarz inequality in \(L^2([0,\infty))\) to the functions \[ x f(x) \qquad\text{and}\qquad f'(x). \] Both vanish outside a bounded interval, so the displayed integrals are finite.
4. Therefore\[ \int_0^\infty f(x)^2\,dx \leq 2 \sqrt{ \int_0^\infty x^2f(x)^2\,dx } \sqrt{ \int_0^\infty f'(x)^2\,dx }. \]
By step 2, \[ \int_0^\infty f^2 = 2\abs{ \int_0^\infty xff' }, \] because the left side is nonnegative. Apply step 3.
5. Q.E.D.
Step 4 is the required inequality.