# Prove $|f(x)-f(a)-df(a)(x-a)|\le \frac{M}{2}\|x-a\|^2$ when $\|d^2f(x)\|$ is bounded.

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:

• Instead of applying the mean value theorem, use the Taylor Expansion of $f$. – WoolierThanThou Feb 27 at 22:41

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 $$|f(x) - f(a) - df(a)(x-a)| \le \frac{M}{2}\|x-a\|^2$$
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, $$f(a+h)=f(a) + Df(a)h + \int^1_0 (1-t)\,\big[D^2f(a+th)\big](h,h)\,dt$$