+- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Analysis (/forumdisplay.php?fid=10) +--- Thread: $f\in\mathscr{C}^2([0,1])\implies |f\,'|\le{\small\displaystyle\int_0^1} (4|f|+|f\,''|)$ (/showthread.php?tid=727) $f\in\mathscr{C}^2([0,1])\implies |f\,'|\le{\small\displaystyle\int_0^1} (4|f|+|f\,''|)$ - elim - 05-04-2018 04:35 PM Proof. Let$\underset{\,}{\,\;}(m,M)=(\min,\max)|f'|{\small([0,1]),\;}|f'|(\alpha)=m,\;|f'|(\beta)=M.$ ${\qquad}$By mean value TH.$,\,f(x)=f(0)+x\cdot f'(\xi)\;(\underset{\,}{\xi}=\theta x,\;0<\theta_x<1).\;\;$ $\underset{\,}{\therefore\quad}\;$If$\;f\ge 0\,$on$\,(0,1),\,$then$\,|f|\ge mx.\,$hence${\small\displaystyle\int_0^1}|f|\ge{\small\displaystyle\int_0^1}mx dx=\large\frac{m}{2}$ $\underset{\,}{\qquad}$If$\,f< 0\,$on$\,(0,1),\,$then$\,-f(1-x)=-f(1)+xf'(\xi)\ge\,mx$ $\underset{\,}{\qquad}$and again we have$\,{\small\displaystyle\int_0^1}|f|={\small -\displaystyle {\int_0^1}}f(1-x)dx\ge\large\frac{m}{2}.$ $\underset{\,}{\qquad}$If$\,f(\eta)=0,\,$for some$\,\eta\in(0,1),\;\;$then$\;|f(x)|=|(x-\eta)f'(\xi)|\;$thus $\underset{\,}{\qquad}{\small\displaystyle{\int_0^1}}|f|\ge{\small\displaystyle{​\int_0^{\eta}}}m(\eta-x)+{\small\displaystyle\int_{\eta}^1}m(x-\eta)=m((\eta-\frac{1}{2})^2+\frac{1}{4})\ge\large\frac{m}{4}.$ $\therefore\;\;\;{\small\displaystyle\int_0^1}(4|f|+|f''|)\ge m+\big|{\small\displaystyle\int_{\alpha}^{\beta}}f''(x)dx\big|\ge{\small m+M-m}\ge |f'(x)|.\;\square$