0
$\begingroup$

I need to find an example of a function $f\in \mathcal{L}_2$, $\lim_{t \to \infty}f(t) \neq 0$. Where the $\mathcal{L}_2$-norm of a signal is defined as $||x(\cdot)||_2^2=\sqrt{\int_0^\infty||x(t)||_2^2dt}$.

However, I feel like this is not possible because the norm needs be bounded, while the limit of the function does not go to 0. Is this a correct observation, or am I missing something? Is there an example that I might have overlooked/missed? If so, what could be such a function that complies to these conditions?

2 Answers 2

0

It doesn't have to satisfy $\lim_{x\to +\infty} f(x)=0$, but "almost always".

I'll explain myself. For example, $f(x)=0$ if $x\notin \mathbb N$ and $f(n)=n$ if $x\in N$ doesn't satisfy $\lim_{x\to +\infty} f(x)=0$, but $\lVert f \rVert_2=0$, because it's equal to the function $g=0$ almost everywhere.

  • 0
    Just one question: I am told by fellows that the mentioned function is discontinous and therefore not integrable. Is this true, or are no condtions for integration violated? – 2017-02-24
  • 0
    It is not true that a discontinuous function is not integrable. Indeed, that one is an example. I don't know which definition of integral you're working with (I assume it's Lebesgue integral), so please let me know. But it is true that if $f$ is continuous and $f\in \mathcal L_2$, then it does have to verify $\lim_{x\to +\infty} f(x)=0$. – 2017-02-24
0

You can produce many such functions with the following "recipe": Choose two sequences, $\{\varepsilon_k\}_k$ and $\{v_k\}$ with $\varepsilon \searrow 0$ and $v_k \to \infty$. Then, set $$f = \sum_{k \in \mathbb{N}} v_k \, \chi_{[k, k + \varepsilon_k]},$$ where $\chi_{[a,b]}$ is the indicator function of an interval. Then, $f$ belongs to $L^2(0,\infty)$ if the series $$\sum_{k \in \mathbb{N}} v_k^2 \, \varepsilon_k$$ converges.

However, since $v_k \to \infty$, the limit $\lim_{x \to \infty} f(x)$ does not exist.