Defining a continuous function $f \in L^1(0,\infty), L^\infty(0,\infty) : f(t) \not\to 0$ as $t \to \infty$

Question

I'm really struggling with this idea, as the function needs to be both continuous and not approach zero at infinity. Without the non-zero limit condition, with $\varepsilon > 0$, I could simply take something like the trapezoidal function $$ f(t) = \left\{ \begin{aligned} 0, && t \notin (1 - \varepsilon,2 + \varepsilon) \\ \frac 1 \varepsilon (t - 1) + 1, &&t \in [1 - \varepsilon, 1]\\ -\frac 1 \varepsilon (t - 2) + 1, &&t \in [2, 2 + \varepsilon]\\ 1, && t \in (1, 2), \end{aligned} \right. $$ which would fulfill the requirements. This is because the function is certainly integrable on the given interval $I = (0,\infty)$, with $$ \Vert f \Vert_1 = \int_I |f(t)| \,\mathrm d t \approx 1 < \infty, $$ and its essential supremum $\Vert f \Vert_\infty = \mathrm{ess\,sup}_{t \in I} |f(t)| = 1$. I guess this question is doing a very good job of undermining my understanding of the subject. Any hints as to how i should approach this?

Answer 1

Put a thinner spike of height $1$ in the interval $(3,4)$. Put a yet thinner spike of height $1$ in the interval $(5,6)$, and so on ad infinitum.

As long as the areas of the spikes form the terms of a convergent series, your function will be integrable.