$ \newcommand{\norm}[1]{\Vert #1 \Vert} \newcommand{\normc}{\Vert \cdot \Vert} $For the three norms, $\normc_1,$ $\normc_2$, $\normc_{\infty}$ on the space of continuous real-valued signals defined on $(0, \infty)$, find a function for which the norm (it could be an improper integral) is finite only for that norm and not for the other two. Show that if $f$ is continuous and $\norm f_1$, and $\norm f_{\infty}$ are both finite then $\norm f_2$ is as well.

The hint in the exercise is that you can define your function piece-wise.

I know that the three norms look like this: $$\norm f_1 = \sum_{t \in T}|f(t)|$$ $$\norm f_2 = \left(\sum_{t \in T}|f(t)|^2\right)^{1/2}$$ $$\norm f_{\infty} = \sup_{t \in T}|f(t)|$$ Where $T$ is the time domain (I believe in this exercise it would be the continuous real-valued signals, but I'm not 100% sure).

I'm not too sure what the functions should look like.

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    $\begingroup$ What's $T$? ${}{}{}$ $\endgroup$ Oct 1, 2020 at 17:43
  • $\begingroup$ Should the last norm in the first paragraph be the sup norm? $\endgroup$ Oct 1, 2020 at 17:58
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    $\begingroup$ @jejove2096 which functions have you tried? Were the norms finite or infinite? $\endgroup$
    – supinf
    Oct 1, 2020 at 18:38
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    $\begingroup$ Sorry what is the time domain? Indeed I am confused since you have real-valued signal defined on $(0,\infty)$, but then your norms seem to say that your signal is defined on a discrete set $T$. $\endgroup$ Oct 1, 2020 at 19:02
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    $\begingroup$ Please don't self-delete your question with a high-quality answer again. The loss of good content, especially the well-written answer, from the network has a negative impact on this site's SEO. $\endgroup$ Jan 15, 2021 at 8:33

1 Answer 1


$ \newcommand{\norm}[1]{\Vert #1 \Vert} \newcommand{\normc}{\Vert \cdot \Vert} $For $p\in\mathbb{R}^{+}$, define

$$\norm{f}_p=\left(\int_0^\infty |f|^pdt\right)^{1/p}$$



(if they exist)

$p=1):$ Consider the function



$$\norm{f}_1=\int_0^\infty |f|dt=\int_0^1\left(\frac{1}{\sqrt{t}}-1\right)dt=1$$

$$\norm{f}_2^2=\int_0^\infty |f|^2dt=\int_0^1\left(\frac{1}{\sqrt{t}}-1\right)^2dt=\int_0^1\left(-\frac{2}{\sqrt{t}}+\frac{1}{t}+1\right)dt=\infty$$


$p=2):$ Consider the function

$$f(t)=\begin{cases} \frac{1}{t^{1/4}} & t\leq 1 \\ \frac{1}{t} & t\geq 1 \end{cases}$$


$$\norm{f}_1=\int_0^\infty |f|dt\geq \int_1^\infty \frac{1}{t}dt=\infty$$

$$\norm{f}_2^2=\int_0^\infty |f|^2dt=\int_0^1\frac{1}{\sqrt{t}}dt+\int_1^\infty\frac{1}{t^2}dt=\frac{5}{3}$$


$p=\infty):$ Consider the function



$$\norm{f}_1=\int_0^\infty |f|dt\geq \int_0^\infty \frac{1}{\sqrt{t+1}}dt=\infty$$

$$\norm{f}_2^2=\int_0^\infty |f|^2dt\geq \int_0^\infty \frac{1}{t}dt=\infty$$


For the second portion, assume that both $\norm{f}_1$ and $\norm{f}_\infty$ exist. From this, we know that $f(t)$ is bounded on $(0,\infty)$ (this bound is $M=\norm{f}_\infty$) and that

$$\norm{f}_1=\int_0^\infty |f|dt<\infty$$

This implies that


(since $\frac{|f|}{M}\leq 1$). Therefore

$$\norm{f}_2^2=\int_0^\infty |f|^2dt=M^2\int_0^\infty \frac{|f|^2}{M^2}dt<M^2\int_0^\infty \frac{|f|}{M}dt=M\norm{f}_1<\infty$$

We conclude $\norm{f}_2$ is also finite.


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