Taylor Theorem inequality

Question

Prove that for all $f\in C^2([0,1])$ with $f(0)=f(1)=0$ and $|f''(x)| \le 1$ $$|f(x)| \le \frac{1}{2}x(1-x)$$ $\forall x \in [0,1]$.

Answer 1

For some $\xi_k\in[0,1]$, $$ f(1)=f(0)+f'(0)+\frac12f''(\xi_1) \quad\text{and}\quad f(0)=f(1)-f'(1)+\frac12f''(\xi_2) $$ implies $$ |f'(0)|\le\frac12 \quad\text{and}\quad |f'(1)|\le\frac12 $$ Let $g(x)=f(x)-\frac12x(1-x)$ and $h(x)=f(x)+\frac12x(1-x)$.

$g(0)=0$, $g'(0)\le0$, and $g''(x)\le0$; therefore, $g(x)\le0$.

$h(0)=0$, $h'(0)\ge0$, and $h''(x)\ge0$; therefore, $h(x)\ge0$.

Thus, $$ \overbrace{-\frac12x(1-x)\le}^{h(x)\ge0}f(x)\overbrace{\le\frac12x(1-x)}^{g(x)\le0} $$

Answer 2

We have, $\displaystyle (1-x)\int_0^x tf''(t)\,dt -x \int_x^1 (t-1)f''(t)\,dt = -f(x)$, for $x \in [0,1]$ (just apply integration by parts on LHS to ease the simplification).

Since, $t\ge 0$ for $t\in [0,1]$ and $(t-1) \le 0$ for $t \in [0,1]$, taking modulus on both sides,

$|f(x)| = \displaystyle \left|(1-x)\int_0^x tf''(t)\,dt -x \int_x^1 (t-1)f''(t)\,dt\right|$

$\le \displaystyle \left|(1-x)\int_0^x tf''(t)\,dt\right| + \left|-x \int_x^1 (t-1)f''(t)\,dt\right|$

$\le \displaystyle \sup\limits_{t\in[0,1]}|f''(t)|. \bigg( (1-x)\int_0^x t\,dt -x \int_x^1 (t-1)\,dt \bigg)$

$= \dfrac{x(1-x)}{2}.\sup\limits_{t\in[0,1]}|f''(t)| \le \dfrac{x(1-x)}{2}$.

Aliter: Define $g(t)=f(t)-\dfrac{t(t-1)}{x(x-1)}f(x)$, on $[0,1]$.

Then, $g(0)=g(x)=g(1)=0$.

Applying Rolle's Theorem twice on $(0,1)$, $\exists \alpha \in (0,1)$ such that $g''(\alpha)=0$.

That is $g''(\alpha) = f''(\alpha) - \dfrac{2}{x(x-1)}f(x)=0$

or, $|f(x)|=\dfrac{x(1-x)}{2}|f''(\alpha)| \le \dfrac{x(1-x)}{2}$.

Answer 3

We want to prove that for each $x$:$$ \exists c\ \ f(x) =\frac 12 f''(c) x(1-x) $$

We want to find such a $c$ via the Rolle theorem (or via the mean value theorem, but we can always go back to the Rolle version).

We can already apply the Rolle theorem, which gives an annulation for $f'$.

Let us modify $f$ to go to $0$ once more: assuming $0<x<1$,

$$ g(u) := f(u) + A_xu(1-u)\\ g(x) = 0\Leftarrow A_x = - \frac{f(x)}{x(1-x)} $$

Now $g(1)=g(x) = g(0)$ hence, applying several times the Rolle theorem:

$$ \exists c \ \ 0=g''(c) = f''(c) - \frac{f(x)}{x(1-x)}(-2)\\ f(x) = -\frac 12 x(1-x)f''(c) $$

NB: the error in the sign does not change the final inequality.