Find an upper bound for $\int_{0}^{1}\cos(x)\frac{x f'(x) - f(x) + f(0)}{x^2}dx$

Question

Let $f : [0, 1] \to \mathbb{R}$ such that $f$, $f'$ and $f''$ are continuous on $[0, 1]$ We want an upper bound for

$$\int_{0}^{1}\cos(x)\frac{x f'(x) - f(x) + f(0)}{x^2}dx$$

which is strictly bounded by $c \cdot \sup_{[0, 1]}|f''(x)|$, for example $c = \frac{3}{2}$. The goal is to prove this.

Currently I just have, using Bonnet's theorem, that

$$\int_{a}^{b} f(x)g(x)dx = B \int_{x_0}^{b}g(x)dx$$

for some $x_0$ in $[a, b]$, given conditions that $f(x) \geq 0$ in $[a, b]$ and $B \geq f(b^-)$

and so in this case $\int_{0}^{1}\cos(x)\frac{x f'(x) - f(x) + f(0)}{x^2}dx = B \int_{x_0}^{1}\frac{x f'(x) - f(x) + f(0)}{x^2}dx$

since $\cos(x) \geq 0$ in $[0, 1]$ and $B \geq f(1^-) = \cos(1)$ for some $x_0$ in $[0, 1]$.

Then we might apply the MVT to get the preceding is equal to $B(1 - x_0)[\frac{c f'(c) - f(c) + f(0)}{c^2}]$ for $c$ in $(0, 1)$. But I am not sure where to proceed from this point. Am I on the right track? Should I be using Bonnet's theorem instead of Taylor approximation or some other technique?

Answer 1

$$I=\int_{0}^{1}\cos(x)\frac{x f'(x) - f(x) + f(0)}{x^2}dx$$

$$|I|\le \int_{0}^{1}\left|\cos(x)\frac{x f'(x) - f(x) + f(0)}{x^2}\right|dx\le \int_{0}^{1}\left|\frac{x f'(x) - f(x) + f(0)}{x^2}\right|dx$$

Taylor expansion,

$$f(x)=f(0)+f'(0)x+\frac{1}2f''(y)x^2,~~~~y\in[0,x]$$

we have

$$\left|\frac{x f'(x) - f(x) + f(0)}{x^2}\right|=\left|\frac{(f'(x) - f'(0))x - \frac{1}2f''(y)x^2}{x^2}\right|=K$$ By Mean Value theorem,

$$f'(x) - f'(0)=f''(z)x,~~~~z\in[0,x]$$ we have

$$K=\left|f''(z)- \frac{1}2f''(y)\right|\le\left|f''(z)\right|+\frac{1}2\left| f''(y)\right|\le \frac{3}2\sup_{[0, 1]}|f''(x)|$$

Therefore,

$$|I|\le \frac{3}2\sup_{[0, 1]}|f''(x)|$$

Answer 2

Taylor's theorem (with $x$ as the “start point“ and $0$ as the “end point”) gives $$ f(0) = f(x) + (0-x)f'(x) + \frac{(0-x)^2}{2}f''(c) = f(x) - xf'(x) + \frac{x^2}{2}f''(c) $$ for some $c$ between $0$ and $x$. It follows that $$ \left| \frac{x f'(x) - f(x) + f(0)}{x^2}\right| = \frac 12 |f''(c)| \le \frac 12\sup_{x \in [0, 1]} |f''(x)| $$ and therefore, since $\cos(x) > 0$ on that interval, $$ \begin{align} \left|\int_{0}^{1}\cos(x)\frac{x f'(x) - f(x) + f(0)}{x^2}\, dx \right| &\le \int_0^1 \cos(x) \, dx \cdot \frac 12\sup_{x \in [0, 1]} |f''(x)| \\ &= \frac {\sin(1)}{2}\cdot \sup_{x \in [0, 1]} |f''(x)| \, , \end{align} $$ i.e. your desired bound $|I| \le c \cdot \sup_{x \in [0, 1]}|f''(x)|$ holds with the factor $$\boxed{c = \frac 12\sin(1) \approx 0.4207} \, .$$

This factor is best possible, equality holds if $f$ is a quadratic polynomial.