# If $M_0$, $M_1$, and $M_2$ are least upper bounds of $|f(x)|$, $|f'(x)|$ and $|f''(x)|$, does $M_1^2\leq 4M_0M_2$ for vector valued functions?

If $a\in\mathbb{R}$, $f\colon(a,\infty)\to\mathbb{R}$ is a twice-differentiable function, and $M_0$, $M_1$, and $M_2$ are least upper bounds of $|f(x)|$, $|f'(x)|$ and $|f''(x)|$, then $M_1^2\leq 4M_0M_2$.

This follows from Taylor's theorem, for taking an interval $(x,x+2h)$ for some $x$ and any $h>0$, there exists some $\xi\in(x,x+2h)$ such that $$f(x+2h)=f(x)+2hf'(x)+2h^2f''(\xi).$$ Rearranging shows $$f'(x)=\frac{1}{2h}(f(x+2h)-f(x))-hf''(\xi)\implies |f'(x)|\leq hM_2+\frac{M_0}{h}.$$ The bound holds if $M_0$ or $M_2$ equals $0$, otherwise taking $h=\sqrt{M_0/M_2}$ gives the bound.

I'm curious, does this bound also work if $\mathbf{f}$ is some vector valued function, $\mathbf{f}(x)=(f_1(x),\dots,f_n(x))$? I tried applying the same argument with Taylor's theorem to each coordinate, in hopes of finding a similar equation $\mathbf{f}(x+2h)=\mathbf{f}(x)+2h\mathbf{f}'(x)+2h^2\mathbf{f}''(\xi)$ but got stuck since the respective $\xi_i$ for each $f_i$ is possibly different. Does the bound still hold, or is there a possible counterexample? Thanks.

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It works. The standard trick is to take the point $t$ where $M_1$ is almost attained, choose the unit vector $u$ so that $\langle f'(t),u\rangle\approx M_1$, and apply the scalar inequality to the function $t\mapsto \langle f(t),u\rangle$
Thanks fedja. Do you mind explaining in more detail how this works? What does the notation $\langle f'(t),u\rangle$ mean, and how do you "apply the scalar inequality" to get the conclusion? –  yunone Jan 12 '12 at 3:31
$\langle\cdot,\cdot\rangle$ is the usual scalar product in $\mathbb R^n$. Note that $\langle f,u\rangle^{(k)}=\langle f^{(k)},u\rangle$, so the scalar function thus obtained has not larger $M_0$, $M_2$ than and essentially the same $M_1$ as the original vector one. –  fedja Jan 12 '12 at 4:01