I'm now stuck on a proof as follows

Suppose $f$ is a smooth function. We define another function $$g(h)=\sup_{x}|f(x+h)-f(x)-f'(x)h-\frac{1}{2}f''(x)h^{2}|$$

Then we have $$g(h)\leq K \min \{h^{2},|h|^{3}\}$$ Where $K$ is a constant depending on $f$ alone.

We have $g(h)\leq Kh^{2}$ for $h$ large and $g(h)\leq K|h|^{3}$ for $h$ near $0$.

The book said we get the conclusion above by using mean value theorems of second and third orders. I don't know how we get that. I think if we use Taylor expansion of $f$ at point $x$, then we get $$f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(x)h^{2}+o(h^{2})$$ Where $o(h^{2})=\frac{f'''(z)}{6}h^{3}$ for some $z \in (x,x+h)$.

I don't understand why $g(h)\leq Kh^{2}$ works well for $h$ large. I think if $h$ is large, then $$\frac{f'''(z)}{6}h^{3}>Kh^{2}$$ for a constant $K$.

Could someone help me with it? Thanks in advance!


1 Answer 1


Can you provide some more details (if any) about how and/or if $f$ and its derivatives are bounded?

The issue here, I believe, is that $z$ in the expression $f'''(z)$ may depend on $h$. An alternative, for large values of $h$, assuming that $f''$ is a bounded function, is to do the following.

$$|f(x+h) - f(x) - f'(x)h - \frac{1}{2}f''(x)h^2| = |f'(x_1)h - f'(x)h - \frac{1}{2}f''(x)h^2|=$$ $$= |hf''(x_2)(x_1-x) - \frac{1}{2}f''(x)h^2|\leq h^2(|f''(x_2)| + \frac{1}{2}|f''(x)|)\leq K_1h^2,$$ where the first two equalities follow from the Mean Value Theorem and the inequality follows from the triangle inequality and $|x_1-x|\leq h$.

On the other hand, for small values of $h$, you may use the Taylor estimate. This time, $z$ depending on $h$ will not cause problems, since to determine $K_2$ so that $g(h)\leq K_2|h|^3$, you may simply take the $\sup$ over some bounded interval of $h$ (again, assuming that $\sup_{x, h} |f'''(z_{x, h})|$ exists).

Finally, it suffices to take $K=\max\{K_1, K_2\}$.


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