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Let $f:[0,\infty)\to\mathbb{R}$ be continuous. I would like to show the following: \begin{equation} \limsup_{t\to\infty}\frac{1}{t}\int_{0}^{t}\lvert f(\tau)\rvert\,\mathrm{d}\tau\leq \limsup_{t\to\infty}\,\lvert f(t)\rvert. \end{equation} In this post, an answer is provided in terms of sequences. Does the same ''strategy'' work here as well?

Following these arguments, we have for $0< s\leq t$, \begin{align} \frac{1}{t}\int_{0}^{t}\lvert f(\tau)\rvert\,\mathrm{d}\tau&= \frac{1}{t}\int_{0}^{s}\lvert f(\tau)\rvert\,\mathrm{d}\tau+ \frac{1}{t}\int_{s}^{t}\lvert f(\tau)\rvert\,\mathrm{d}\tau\\ &\leq\frac{1}{t}\int_{0}^{s}\lvert f(\tau)\rvert\,\mathrm{d}\tau+ \frac{t-s}{t}\sup_{\tau\in[s,t]}\lvert f(\tau)\rvert\\ &\leq\frac{1}{t}\int_{0}^{s}\lvert f(\tau)\rvert\,\mathrm{d}\tau+ \frac{t-s}{t}\sup_{\tau\geq s}\,\lvert f(\tau)\rvert. \end{align} Taking the upper limit as $t\to\infty$, we get \begin{equation} \limsup_{t\to\infty}\frac{1}{t}\int_{0}^{t}\lvert f(\tau)\rvert\,\mathrm{d}\tau\leq \sup_{\tau\geq s}\,\lvert f(t)\rvert. \end{equation} Do I now simply pass to the limit as $s\to\infty$ or is there any ''trap'' I have not thought over?

User129599
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There is no trap. You proved that for all $s>0$ the inequality $$\limsup_{t \to \infty} \frac{1}{t} \int_0^t |f( \tau)| \ \mathrm d \tau \le \sup_{\tau \ge s} |f( \tau)|$$ holds. In particular, taking the infimum of RHS you get $$\limsup_{t \to \infty} \frac{1}{t} \int_0^t |f( \tau)| \ \mathrm d \tau \le \inf_{s>0} \left( \sup_{\tau \ge s} |f( \tau)| \right)$$ Which is exactly the inequality you needed: $$\limsup_{t \to \infty} \frac{1}{t} \int_0^t |f( \tau)| \ \mathrm d \tau \le \limsup_{s \to \infty} |f( s)| $$

Crostul
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