[Home Squat University Mathematics Magazine] NO. 413 phase interpolation inequalities

Set $$\bex k\geq 2,\quad f\in c^k (\bbr), \quad m_j=\sup_{x\in\bbr}|f^{(j)} (x) |\ (j=0,1,\cdots,k). \eex$$ $$\bex M_j\leq 2^\frac{j (k-j)}{2}m_0^{1-\frac{j}{k}}m_k^\frac{j}{k}\ (j=0,1,\cdots,k). \eex$$

Prove:

(1). Only the conclusion of the $0<j<k$ proof shall be established.

(2). Toward the $k $ as a mathematical induction method. When $k =2$, to $j =1$, by $$\bex F (x+h) =f (x) +f ' (x) h+\frac{f ' (\xi)}{2}h^2, \eex$$ $$\bex f (x-h) =f (x)-F ' (x) h+\frac{f "(\eta)} {2}h^2. \eex$$ subtract and have $$\bex |f ' (x) |\cdot 2h \leq 2m_0+m_2h^2\ra |f ' (x) |\leq \frac{m_0}{h}+\frac{m_2h}{2}. \eex$$ $$\bex H=\sqrt{\frac{2m_0}{m_2}}\lra \frac{m_0}{h}=\frac{m_2h}{2}, \eex$$ $$\bex |f ' (x) |\leq 2\cdot \frac{M_2 }{2}\sqrt{\frac{2m_0}{m_2}}=\sqrt{2m_0m_2}. \eex$$ hypothesis Conclusion when the $k =n$ is established, when the $k =n+1$, to $0<j<n+1$, $$\beex \bea m_j&\leq 2^\frac{j (n-j)}{2}m_0^{1-\frac{j}{n}}m_ n^\frac{j}{n}\\ &\leq 2^\frac{j (n-j)}{2}m_0^{1-\frac{j}{n}}\sex{2^\frac{n}{2} M_0^{1-\frac{n}{n+1}}M_{n+1}^\ frac{n}{n+1}}^\frac{j}{n}\\ &=2^\frac{j (n+1-j)}{2} m_0^{1-\frac{j}{n+1}}m_{n+1}^\frac{j}{n+1}. \eea \eeex$$

[Home Squat University Mathematics Magazine] NO. 413 phase interpolation inequalities