108學年度｜微積分模組班｜Class 01｜Note 3.

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Note 3: Concave and Strictly concave

  • Date: 108/10/21 (Weel 07)
  • Place: 新生教學館 301

Part 1: Concave v.s. Strictly Concave

Problem 1.1
  • Suppose that $f(x)$ and $f'(x)$ are differentiable on $(a, b)$.
  1. If $f''(x) \geqslant 0$, for all $x \in (a, b)$, then $f\left(\dfrac{x_1+x_2}{2}\right) \leqslant \dfrac{f(x_1)+f(x_2)}{2}$, for all $x_1 < x_2 \in (a,b)$.
  2. If $f''(x) > 0$, for all $x \in (a, b)$, then $f\left(\dfrac{x_1+x_2}{2}\right) < \dfrac{f(x_1)+f(x_2)}{2}$, for all $x_1 < x_2 \in (a,b)$.
  3. Is it true that if $f\left(\dfrac{x_1+x_2}{2}\right) \leqslant \dfrac{f(x_1)+f(x_2)}{2}$ for all $x_1 < x_2 \in (a,b)$ then $f''(x) \geqslant 0$ ?
  4. Is it true that if $f\left(\dfrac{x_1+x_2}{2}\right) < \dfrac{f(x_1)+f(x_2)}{2}$ for all $x_1 < x_2 \in (a,b)$ then $f''(x) > 0$?
Solution.
  1. By direct computation:\begin{align*} & \dfrac{f(x_1)+f(x_2)}{2} - f\left( \dfrac{x_1 + x_2}{2} \right) \\ = \hspace{0.2cm}& \dfrac{1}{2}\left[f(x_1) - f\left( \dfrac{x_1 + x_2}{2} \right) \right] + \dfrac{1}{2}\left[f(x_2) - f\left( \dfrac{x_1 + x_2}{2} \right) \right] \\ \overset{(1)}{=} \hspace{0.2cm}& \dfrac{1}{2}f'(\xi_1)\left(x_1 - \dfrac{x_1+x_2}{2}\right) + \dfrac{1}{2}f'(\xi_2)\left(x_2 - \dfrac{x_1 + x_2}{2}\right) \\ = \hspace{0.2cm}& \dfrac{1}{4}f'(\xi_1)(x_1-x_2) + \dfrac{1}{4}f'(\xi_2)(x_2-x_1) \\ = \hspace{0.2cm}& \dfrac{1}{4}(x_2 - x_1)(f'(\xi_2) - f'(\xi_1)) \\ \overset{(2)}{=} \hspace{0.2cm}& \dfrac{1}{4}(x_2 - x_1) \cdot f''(\eta) (\xi_2 - \xi_1) \end{align*}
    2. Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, we may apply M.V.T, and hence equality (1) do holds for some $\xi_1 \in \left(x_1, \dfrac{x_1+x_2}{2}\right)$ and $\xi_2 \in \left(\dfrac{x_1 + x_2}{2}, x_2\right)$. Similarly, since $f'$ is continuous on $[a,b]$ and differentiable on $(a,b)$, we may apply M.V.T, and equality (2) holds for some $\eta \in (\xi_1, \xi_2)$. Since $x_2 > x_1$, $\xi_2 > \xi_1$ and $f'(\eta) \geqslant 0$ by assumption, we have: $$\dfrac{f(x_1)+f(x_2)}{2} - f\left( \dfrac{x_1 + x_2}{2} \right) = \dfrac{1}{4}(x_2 - x_1) \cdot f''(\eta) (\xi_2 - \xi_1) \geqslant 0$$, and hence the result.
  2. Similar as (a).
  3. True, we will prove this directly. 
    4. Let $x_1=y=x-h,x_2=x+h$ for some $x \in (a,b)$, $h>0$ and $a \leqslant x_1=y < x < x_2 \leqslant b$. By the assumption, we have: \begin{align*} f\left(\dfrac{x_1+x_2}{2}\right) \leqslant \frac{f(x_1)+f(x_2)}{2} & \Longrightarrow f(x) \leqslant \frac{1}{2}(f(x-h)+f(x+h)) \\ & \Longrightarrow f(x+h))-2f(x)+f(x-h) \geqslant 0 \\ & \Longrightarrow (f(x+h))-f(x))-(f(x))-f(x-h)) \geqslant 0 \\ & \Longrightarrow \frac{f(x+h)-f(x)}{h}-\frac{f(y+h)-f(y)}{h} \geqslant 0 \\ & \Longrightarrow \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}-\frac{f(y+h)-f(y)}{h} \geqslant 0 \\ & \Longrightarrow f'(x)-f'(y) \geqslant 0, \forall a \leqslant y < x \leqslant b \end{align*}
    Hence, $f'(x)$ is increasing on $(a,b)$, that is, $f''(x) \geqslant$ on $(a,b)$.
  4. False, let $f(x)=x^4$, we will prove that though $f''(x)=12x^3 \geqslant 0$ and actually $f''(0)=0$, but $ f\left(\dfrac{x_1+x_2}{2}\right) > \dfrac{f(x_1)+f(x_2)}{2}$, for all $x_1 < x_2 \in \mathbb{R}$. Let $g(x)=x^2$, notice $g''(x)=2>0$, for all $x \in \mathbb{R}$, by the result in (b), we have that $g\left(\dfrac{x_1+x_2}{2}\right) > \dfrac{g(x_1)+g(x_2)}{2}$, for all $x_1 < x_2 \in \mathbb{R}$. Hence : \begin{align*} f\left(\dfrac{x_1+x_2}{2}\right) = \left(\frac{x_1+x_2}{2}\right)^4 &= \left[\left(\dfrac{x_1+x_2}{2}\right)^2\right]^2 \\ &= \left[g\left(\dfrac{x_1+x_2}{2}\right)\right]^2 \\ &>\left[\dfrac{(g(x_1)+g(x_2)}{2}\right]^2 \\ &=g\left(\dfrac{x_1^2+x_2^2}{2}\right) \\ &>\dfrac{g(x_1^2)+g(x_2^2)}{2} \\ &=\dfrac{x_1^4+x_2^4}{2} =\dfrac{f(x_1)+f(x_2)}{2} \end{align*}
    5. Hence $f\left(\dfrac{x_1+x_2}{2}\right)>\dfrac{f(x_1)+f(x_2)}{2}$, for all $x_1 < x_2 \in \mathbb{R}$, but $f''(x)$ isn't strictly positive.

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