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Problem: Suppose $f:\Bbb R\to\Bbb R$ is a non-negative uniformly continuous function and $\displaystyle\int_{-\infty}^\infty f(x)\ dx <+\infty$. Prove $f$ is bounded.

I got this problem in my final exam today but couldn't figure it out. Any hint/solution is welcome.

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    www.maths.tcd.ie/pub/ims/bull53/R5301.pdf2012-11-30

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Hint: take $\delta > 0$ so $|x - y| < \delta$ implies $|f(x) - f(y)| < 1$. If $f(x) > n$ then $\int_{x-\delta}^{x+\delta} f(y)\ dy > \ldots$.

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Proceed by contradiction. Pick $A$ and $\varepsilon$ with $A > 2\varepsilon$. If the function $f$ is not bounded, then there exists a sequence $x_n \rightarrow \infty$ such that $f(x_n) > A$. Since $\int_{-\infty}^{\infty} f(x) dx < \infty$ there exists a sequence of $h_n \rightarrow 0$ such that $f(x_n + h_n) < A/2$ (the function $f$ cannot remain large for too long). Hence $f(x_n + h_n) - f(x_n) \geq A/2$. But once $n$ is large enough we will have $f(x_n + h_n) - f(x_n) < \varepsilon$ by uniform continuity, because $h_n \rightarrow 0$. Hence $A/2 \leq f(x_n + h_n) - f(x_n) < \varepsilon$, a contradiction with $A > 2 \varepsilon$.