If $f(t)$ fulfill $\int_{-\infty}^\infty |f(t)|dt < \infty$, Is then true $\int_{-\infty}^\infty |f(t)|^2 dt < \infty$? If not, What else is needed?

Question

If $f(t)$ is a continuous time-limited function such that fulfills $\int_{-\infty}^\infty |f(t)|dt < \infty$, Is then true that $\int_{-\infty}^\infty |f(t)|^2 dt < \infty$? If not, What else is needed?

I am working into this question looking for the conditions that will makes time-limited functions (so of unlimited bandwidth), to have a maximum rate of change $\max_t |df(t)/dt|$.

I think that imposing the restriction of $f(t)$ being of bounded total variation is too "strong" since it is contained in its definition $\int_{-\infty}^\infty |f'(t)|dt < \infty \Rightarrow f'(t)\,\text{bounded}$, and also I have the problem of the discontinuity of the derivative in the domain boundaries $\partial t = \{t_0,\,t_F\}$ so the total variation has to be defined within the compact support as $\int_{t_0}^{t_F} |f'(t)|dt < \infty \Rightarrow f'(t)\,\text{bounded} \in (t_0,\,t_F)$.

Working in the question I found that at least any function $f(t)$ which fulfill that: $$\int_{-\infty}^{\infty}|X(w)|dw < \infty$$ will have a bounded maximum slew rate $\max_t |df(t)/dt|$, whith: $$ X(w) = jwF(w) + e^{-iwt_F}f(t_F)-e^{-iwt_0}f(t_0)$$ and $F(w) = \int_{t_0}^{t_F} f(t)e^{-iwt}dt$ the Fourier Transform of time-limited function $f(t)$, and the other terms are to avoid the effects of the discontinuities at the boundaries $\partial t$ in the frequencies domain.

If the function $f(t)$ is of bounded energy $$ E° = \int_{-\infty}^\infty |f(t)|^2 dt = \int_{t_0}^{t_F} |f(t)|^2 dt = \int_{-\infty}^\infty |F(w)|^2 dw < \infty$$ through the Parseval's Theorem, I would like to relate them as "since is squared-integrable, then its maximum slew rate is bounded", but I don´t believe that directly $\int_{-\infty}^\infty |X(w)|^2dw < \infty \Rightarrow \int_{-\infty}^\infty |X(w)|dw < \infty$, so I am looking for which is the "least restricted additional condition" that will leads to a bounded maximum rate of change.

Also, in this paper I found that a integral $\int_{a>0}^\infty f(t)dt < \infty$ if $\lim\limits_{x \to \infty} (x^kf(x)) = A_k < \infty,\,k>1$, but I don't know if it can be related to what I want to do.

Answer 1

A counterexample:

Let $\{I_{n}\}$ be a disjoint sequence of intervals of length $\dfrac{3}{n^{3}}$.

We divide each $I_{n}$ into there parts. In the middle part we let $f(x)$ to be $n$, while in the left and right parts of $I_{n}$, $f(x)$ are the splines joining to the $x$-axis, and $f(x)$ are zero elsewhere, so this $f$ is continuous and nonnegative.

The integral $\displaystyle\int f(x)dx$ can be controlled by $\displaystyle\sum_{n}\dfrac{3n}{n^{3}}=3\sum_{n}\dfrac{1}{n^{2}}<\infty$.

But then the integral $\displaystyle\int f(x)^{2}dx\geq\sum_{n}\dfrac{n^{2}}{n^{3}}=\sum_{n}\dfrac{1}{n}=\infty$.

Answer 2

When I made the question some confusion was created because of describing the function as time-limited, when they are actually labeled as finite-duration functions.

As other have answer in the comments, $\int f^2 \leq \|f\|_\infty \, \int |f|$ so in general the answer is that not only absolute integrability is required, the function $f$ also must be bounded.

But what was tacitly implied in the question, noting my first paragraph here, is that the function $f(t)$ I am asking is a scalar one-variable function of finite-duration, so the function $f(t)$ is of compact-support in the time domain (I am not sure if this still stands if the function $f$ where of more than one variable, as example $f(x,t)$ will be limited only in the variable $t$ but in not in $x$, then probably the following explanation cannot holds), and since there is a theorem that states that continuous and compact supported functions are bounded, for the specific function $f(t)$ of the main question, it is already stated that fulfill $\|f(t)\|_\infty < \infty$, so being $f(t)$ absolute integrable was sufficient conditions in this case to be energy limited:

scalar one-variable continuous finite-duration functions that are absolute integrable, are energy limited.