An inequality with integrals Let $\phi\colon[0,1] \to \mathbb R$ be such that $\phi,\phi^\prime,\phi^{\prime\prime}$ are continuous on $[0,1]$, then the following inequality holds:
$$\int_0^1\cos x\frac{x\phi^\prime(x)-\phi(x)+\phi(0)}{x^2}\mathrm dx < \frac32\|\phi^{\prime\prime}\|_\infty$$
I have no idea how to solve this problem, could you help me please?
 A: We have 
\begin{align*}
x\phi'(x)-\phi(x)+\phi(0)&=x\phi'(x)-\int_0^x\phi'(t)dt\\
&=\int_0^x\left(\phi'(x)-\phi'(t)dt\right)\\
&=\int_0^x\int_t^x\phi''(s)dsdt, 
\end{align*}
hence 
$$\left|\frac{x\phi'(x)-\phi(x)+\phi(0)}{x^2}\right|\leq \frac{\lVert \phi''\rVert}{x^2}\int_0^x\int_t^xdsdt=\frac{\lVert \phi''\rVert}{x^2}\int_0^x(x-t)dt=\frac{\lVert \phi''\rVert}2.$$
Therefore, the integral is convergent and
$$\int_0^1\cos x\frac{x\phi'(x)-\phi(x)+\phi(0)}{x^2}dx\leq\frac{\lVert \phi''\rVert}2 \int_0^1\cos xdx =\frac{\lVert \phi''\rVert}2\sin 1\leq \frac{\lVert \phi''\rVert}2,$$
unless I'm misunderstanding something.
A: Use Taylor series twice (I write f for $\phi$ cause I'm lazy):
$f(x) = f(0) + xf'(0) + \frac{x^2}{2} f''(c)$
and
$f'(x) = f'(0) + xf''(d)$
where $0 \le c, d \le 1$.
From the second one,
$f'(0) = f'(x) - xf''(d)$.
Putting this in the first,
$f(x) = f(0) + x(f'(x)-xf''(d)) + \frac{x^2}{2} f''(c)$
so
$x f'(x) - f(x) + f(0) = -x^2 f''(d) + \frac{x^2}{2}f''(c)$
or
$$\frac{x f'(x) - f(x) + f(0)}{x^2} = -f''(d)+f''(c)/2$$.
Putting this in the integral, since $|\cos| \le 1$ gives the result.
It will be interesting to see how much this agrees with the answer entered while I was entering this
