Let $f:[0, +\infty) \rightarrow \mathbb{R}$ be an integrable and uniformly continuous function. Prove that $\displaystyle \lim_{x \rightarrow +\infty} f(x)=0$.
(Berkley Examinations)
Solution
Suppose on the contrary that the limit is not zero. Then we can pick an $a$ and a sequence $x_n$ such that it holds $x_n \geq x_{n_1} +1$ as well as $f(x_n)>a \;\;\; (*)$. Since $f$ is uniformly continuous there exists $\delta>0$ such that $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\dfrac{a}{2} \;\;\;\; (**)$.
However in the interval $\displaystyle I_n = \left [ x_n -\frac{\delta}{2}, x_n + \frac{\delta}{2} \right ]$ holds $\displaystyle \left | \int_{I_n}f(x)\, {\rm d}x \right |> \frac{a\delta}{2}$ contradicting the integrability of $f$. (this follows as a sequence of $(*), \; (**)$.)
Hence the limit is zero and the exercise comes to an end.
(Berkley Examinations)
Solution
Suppose on the contrary that the limit is not zero. Then we can pick an $a$ and a sequence $x_n$ such that it holds $x_n \geq x_{n_1} +1$ as well as $f(x_n)>a \;\;\; (*)$. Since $f$ is uniformly continuous there exists $\delta>0$ such that $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\dfrac{a}{2} \;\;\;\; (**)$.
However in the interval $\displaystyle I_n = \left [ x_n -\frac{\delta}{2}, x_n + \frac{\delta}{2} \right ]$ holds $\displaystyle \left | \int_{I_n}f(x)\, {\rm d}x \right |> \frac{a\delta}{2}$ contradicting the integrability of $f$. (this follows as a sequence of $(*), \; (**)$.)
Hence the limit is zero and the exercise comes to an end.
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