Let \(\phi\) be a real-valued function defined on \([0,1]\) such that \(\phi\), \(\phi'\), and \(\phi''\) are continuous on \([0,1]\). Prove that \[\int_0^1 \cos x \frac{x\phi'(x)-\phi(x)+\phi(0)}{x^2}\,dx<\frac{3}{2}||\phi''||_\infty,\] where \(||\phi''||_\infty = \sup_{[0,1]}|\phi''(x)|.\) Note that \(3/2\) may not be the smallest possible constant.
Solution
1. Rewrite the numerator as an integral.
let \(N(x) \da x\phi'(x) - \phi(x) + \phi(0)\). Differentiating, \[ \frac{d}{dx}\big(x\phi'(x) - \phi(x) + \phi(0)\big) = \phi'(x) + x\phi''(x) - \phi'(x) = x\phi''(x), \] and \(N(0) = 0\cdot\phi'(0) - \phi(0) + \phi(0) = 0\). Hence by the fundamental theorem of calculus, \[ N(x) = \int_0^x y\,\phi''(y)\,dy . \]
2. Bound \(|N(x)|\).
by 1, \[ |N(x)| \le \int_0^x y\,|\phi''(y)|\,dy \le \norm{\phi''}_\infty \int_0^x y\,dy = \norm{\phi''}_\infty \frac{x^2}{2} . \]
3. Bound the integrand.
2. (The integrand is bounded as \(x \to 0^+\) by the same estimate, so it extends continuously to \(0\).)
4. Conclude.
integrating 3 over \([0,1]\), \[ \int_0^1 \cos x \,\frac{x\phi'(x) - \phi(x) + \phi(0)}{x^2}\,dx \le \int_0^1 \frac{\norm{\phi''}_\infty}{2}\,dx = \frac{1}{2}\norm{\phi''}_\infty < \frac{3}{2}\norm{\phi''}_\infty . \] (The constant \(3/2\) in the problem is not optimal; \(1/2\) suffices.)
5. Q.E.D.