Application of Taylor's theorem, Baby Rudin 5.15

Question

Suppose a ∈ R1, f is a twice-differentiable real function on (a,∞), and M0, M1, M2 are the least upper bounds of $|f(x)|, |f'(x)|, |f''(x)|,$ respectively, on $(a, ∞)$.

Prove that $M_1^2 ≤ 4M_0M_2$.

Hint: If $h > 0,$ Taylor’s theorem shows that $f'(x) = \frac{1}{2h} [f(x + 2h) − f(x)] − hf''(ξ)$ for some $ξ ∈ (x, x + 2h). $ . . .

My question is how to apply Taylor's theorem so that we get $$f'(x) = \frac{1}{2h} [f(x + 2h) − f(x)] − hf''(ξ)$$ for some $ξ ∈ (x, x + 2h).$

I use $x$ and $y$ in place of $\alpha$ and $\beta$ -- easier to type. — RRL, Apr 10 '19 at 06:27
Correct Taylor: $f(\beta) = f(\alpha) + f'(\alpha)(\beta - \alpha) + \ldots$, and $\alpha$ and $\beta$ are any points in the domain where you want to bound $f'$. — RRL, Apr 10 '19 at 06:31
Possible duplicate of Proving that $(\sup_{x\in R}|f'(x)|)^2\leq 4\sup_{x\in R}|f(x)|\cdot\sup_{x\in R}|f''(x)|$. — Martin R, Apr 10 '19 at 07:32
Another one: https://math.stackexchange.com/q/326684 – all found with Approach0 — Martin R, Apr 10 '19 at 07:34

RRL · Accepted Answer · 2019-04-10T06:34:49.467

2

For any $x$ and $y$, $f(y) = f(x) + f'(x)(y-x) + \frac{1}{2} f''(\xi)(y-x)^2,$ is the Taylor approximation where $\xi$ is some point in between $x$ and $y$.

Thus,

$$f'(x)(y-x) = f(y) - f(x) - \frac{1}{2}f''(\xi)(y-x)^2.$$

If $|f(x)| \leqslant M_0$ and $|f''(x)| \leqslant M_2$ are bounds, then

$$|f'(x)| \leqslant \frac{|f(x)| + |f(y)|}{|y-x|} + \frac{1}{2} |f''(\xi)| |y-x| \leqslant \frac{2M_0}{|x-y|}+ \frac{M_2}{2}|x-y|.$$

Now notice that

$$\min_{0 < z < \infty} \left(\frac{2M_0}{z} + \frac{M_2z}{2}\right)= 2 \sqrt{M_0M_2}$$

You should be able to finish from here.

edited Apr 10 '19 at 06:34

answered Apr 10 '19 at 06:21

RRL

90,707

How do you know that is the minima? – Peter_Pan Apr 10 '19 at 06:37
$|f'(x)|$ is less than $2M_0/|x-y| + M_2|x-y|/2$ no matter what we pick for $y$. – RRL Apr 10 '19 at 06:41
So $|x-y|$ can range in $(0,\infty)$ and $|f'(x)|$ must be less than the greatest lower bound. – RRL Apr 10 '19 at 06:42
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Solve for the minimum of that expression $(2M_0/z + M_2z/2)$ by taking the derivative and setting it equal to $0$. The minimum is at $z = 2\sqrt{M_2/M_0}$. – RRL Apr 10 '19 at 06:46
I now understand that, but does it matter that it is the min? Could we just choose $|x−y| = \sqrt { 2M_2/M_0} $? – Peter_Pan Apr 10 '19 at 06:57
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We could put any positive value in for $|y-x|$ and it would be an upper bound for all $|f’(x)|$ but we didn’t know that until we proved it. Also we now know that the specific value you choose in your comment gives us the least upper bound for the derivative. – RRL Apr 10 '19 at 07:31

Application of Taylor's theorem, Baby Rudin 5.15

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