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Suppose $a\in \mathbb{R}^p$ and $f$ is a real-valued function whose second-order partial derivatives all exist and are continuous on $B_r(a)$. Also, suppose that the operator norm $\|d^2f(x)\|$ of the matrix $d^2f(x)$ is bounded by $M$ on $B_r(a)$. Prove that $$|f(x)-f(a)-df(a)(x-a)|\le \frac{M}{2}\|x-a\|^2$$ for all $x\in B_r(a)$.

If I divide both sides by $\|x-a\|$, I get $$\frac{|f(x)-f(a)-df(a)(x-a)|}{\|x-a\|}\le \frac{M}{2}\|x-a\|.$$

What I know:

  • The left side goes to $0$ as $x\to a$
  • As a corollary to the Mean Value Theorem, $|df(x)-df(a)|\le M\|x-a\|$
  • From the Mean Value Theorem, $f(x)-f(a) = df(c_1)(x-a)$ for some $c_1$ on the line segment from $x$ to $a$
  • Less certain about this, but $df(x)-df(a)=d^2f(c_2)(x-a)$ for some $c_2$ on the line segment from $x$ to $a$

It seems like I have all the pieces I need, but I'm not sure how to put them together.

I imagine the $\frac{M}{2}$ comes from having $a$ be the center of the ball. The statements above hold even if $a$ were another arbitrary point in the ball, so it makes intuitive sense that we are bounded by $\frac{M}{2}$ because we only have half the distance to work with. But I'm not sure how to put this all together.


Possibly relevant:

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    $\begingroup$ Instead of applying the mean value theorem, use the Taylor Expansion of $f$. $\endgroup$ – WoolierThanThou Feb 27 at 22:41
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Following WoolierThanThou's suggestion:

The Taylor expansion of $f$ is $$f(x)=f(a) + df(a)(x-a) + \frac{1}{2}d^2f(c)(x-a)^2$$ for some $c$ on the line segment from $a$ to $x$.

Equivalently, $$f(x) - f(a) - df(a)(x-a) = \frac{1}{2}d^2f(c)(x-a)^2$$

Notice $(x-a)^2 = \|x-a\|^2$. Because $B_r(a)$ is convex, $c \in B_r(a)$. Thus $\|d^2f(c)\| \le M$. So we have $$|f(x) - f(a) - df(a)(x-a)| \le \frac{M}{2}\|x-a\|^2$$

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Suggestion: use Taylor's theorem in several variables (See for example Serge Lang's book on Analysis of Multuvariate Calculus). For all $h$ small enough, $$ f(a+h)=f(a) + Df(a)h + \int^1_0 (1-t)\,\big[D^2f(a+th)\big](h,h)\,dt $$

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