Exploring Taylor's Theorem to Prove an Inequality

[laleler1]

Homework Statement

$f(x)$ 2 times differentiable function on $(0, \infty)$, and $\lim\limits_{x \rightarrow \infty} f(x)=0$. there is a $M$ such that $M=\sup\limits_{x>0}\vert f^{\prime\prime}(x) \vert$. And also for $L>0$

$g(L)=\sup\limits_{x>L}\vert f(x) \vert$, and $h(L)=\sup\limits_{x>L}\vert f^{\prime}(x) \vert$.

Then, for and $\delta > 0$, show that holds

$h(L) \leq \frac{2}{\delta}g(L) + \frac{\delta}{2}M$

Homework Equations

The Attempt at a Solution

I don't know how should I start,
but I think we need to use Taylors theorem,
can you help...

 

[mighty2000]

Hi there,

Yes, you are correct in thinking that Taylor's theorem may be useful in solving this problem. Let's start by considering the definition of the derivatives of f(x):

f'(x) = \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}

f''(x) = \lim\limits_{h \rightarrow 0} \frac{f'(x+h)-f'(x)}{h}

Now, since we know that f(x) is twice differentiable on (0, \infty) and that \lim\limits_{x \rightarrow \infty} f(x)=0, we can apply Taylor's theorem to f(x) and f'(x) to obtain the following:

f(x+h) = f(x) + hf'(x) + \frac{h^2}{2}f''(x+\theta h)

f'(x+h) = f'(x) + hf''(x+\theta h)

Where \theta is some value between 0 and 1. Now, if we take the absolute value of these equations and use the fact that f(x) and f'(x) are both bounded on (0, \infty), we can see that:

\vert f(x+h) \vert \leq \vert f(x) \vert + \vert hf'(x) \vert + \frac{h^2}{2}\vert f''(x+\theta h) \vert

\vert f'(x+h) \vert \leq \vert f'(x) \vert + \vert hf''(x+\theta h) \vert

Now, let's define a new function g(x) = \vert f(x) \vert and use the definition of g(L) given in the problem statement. We can then rewrite the first equation as:

g(x+h) \leq g(x) + \vert hf'(x) \vert + \frac{h^2}{2}\vert f''(x+\theta h) \vert

And similarly, for the second equation:

\vert f'(x+h) \vert \leq \vert f'(x) \vert + \vert hf''(x+\theta h) \vert

Now, let's take the supremum of both sides of these inequalities over the interval x>L, where L>0. This gives us:

g(x