Maximum

Real Analysis
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Papapetros Vaggelis
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Posts: 426
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Maximum

#1

Post by Papapetros Vaggelis »

Let \(\displaystyle{f:\left[0,+\infty\right)\longrightarrow \mathbb{R}}\) be a continuous function such that

\(\displaystyle{f(0)=2}\) and \(\displaystyle{\lim_{x\to +\infty}f(x)=0}\).

Prove that the function \(\displaystyle{f}\) has a maximum value at \(\displaystyle{\left[0,+\infty\right)}\) .
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Riemann
Posts: 178
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: Maximum

#2

Post by Riemann »

Hello Vaggelis. Here is a solution:

The second condition tells us that there exists an $M>0$ such that for every $ x> M$ to hold $f(x)<2$. Since $f$ is continuous on $[0, M]$ a maximum is attained and let that be at $f(x_0)$. Then:

$$f(x_0) \geq f(x) \;\; \forall x \in [0, M] \quad \quad \text{and} \quad \quad f(x_0) \geq f(0)= 2 > f(x) \; \; \forall x \in [M, +\infty)$$

Hence $f(x_0)$ is maximum of $f$.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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