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Rudin PMA p.116

Let $f\colon (a,\infty) \rightarrow \mathbb{R}^k$ be a twice differentiable function $ (a\in \mathbb{R})$.

Suppose $|f|,|f'|,|f''|$ has finite upper bounds $M_0,M_1,M_2$ respectively.

How do i prove that ${M_1}^2 \leq 4M_0 M_2$?

I have proved, if $f$ is a real function, above inequality holds.

I have found that, for an arbitrary $h>0$ and $x\in (a,\infty)$,

$f(x+2h)=f(x)+2hf'(x)+ 2h^2 ({f_1}''(\xi_1), ... , {f_k}''(\xi_k))$ where $\xi_i \in (x,x+2h)$.

Does this imply the inequality above?

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    So far, I have proved that ${M_1}^2≦4M_0 M_2 k$2012-11-20

1 Answers 1

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Here is one way to use the fact for real functions in order to get it for vector valued functions.

Let $f = \langle f_1, \cdots, f_n \rangle$ and pick some $y\in (a,\infty)$. Then define $c = \langle f_1'(y), \cdots, f_n'(y) \rangle$.

Then $c \cdot f$ is real valued so we can apply our result to it, which gives $ \lvert (c \cdot f)'(y) \rvert^2 \leq 4 \sup \lvert c \cdot f\rvert\sup\lvert (c\cdot f)''\rvert$ Since $c$ is a constant the RHS is $4 \lvert c \rvert ^2 M_0 M_2$, where $M_0$ and $M_2$ are as you have defined them, and the LHS is exactly $\lvert c\rvert^4$. So we have $ \lvert c \rvert^4 \leq 4 \lvert c \rvert ^2 M_0M_2.$ And since this is true for every choice of $y$ it's also true for the supremum, so $ M_1^4 \leq 4 M_1^2 M_0 M_2.$