# Proving M_1^2 ≤ 4M_0M_2 in Real Analysis

• ehrenfest
In summary, the conversation discusses the proof of M_1^2 <= 4M_0M_2 for a twice-differentiable real function on (a, infinity) with given least upper bounds for its absolute value, first derivative, and second derivative. The conversation mentions using Taylor's Theorem and a hint provided in Rudin's textbook to show this inequality.

## Homework Statement

Suppose $a \in \mathbb{R}$, f is a twice-differentiable real function on (a, \infinty) and M_0,M_1,M_2 are the least upper bound of $|f(x)|,|f'(x)|,|f''(x)|$, respectively on (a,\infinity). Prove that

$$M_1^2\leq 4 M_0 M_2$$

## The Attempt at a Solution

That is equivalent to showing that M_2 x^2 +M_1 x +M_0=0 has a real solution.

I was trying to use Taylor's Theorem which says that if \alpha and \beta are distinct points in (a,\infinity) then there exists x between \alpha and \beta that makes the following equation true:

$$f(\beta) = f(\alpha) + f'(\alpha)(\beta-\alpha) + f''(x) (\beta-\alpha)^2/2$$

I could take the absolute value of both sides and then use triangle inequality but I did not see how to get anywhere with that.

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Did you look at the hint given in rudin?

How was I supposed to know there was a hint on the next page? That is a serious question.

OK. The hint says:

If h>0, Taylor's theorem show that

$$f'(x) = \frac{1}{2h}[f(x+2h)-f(x)]-hf''(\xi)$$

for some $\xi \in (x,x+2h)$. Hence

$$|f'(x)| \leq hM_2+M_0/h$$

I don't see how the last inequality is useful (even if you square it).

Square it and remember that it holds for ANY h > 0. There is a particular choice of h that will yield the result.