# Functional Analysis: Finding finite functions over different norms

$$\newcommand{\norm}{\Vert #1 \Vert} \newcommand{\normc}{\Vert \cdot \Vert}$$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.

• What's $T$? ${}{}{}$ Oct 1 '20 at 17:43
• Should the last norm in the first paragraph be the sup norm? Oct 1 '20 at 17:58
• @jejove2096 which functions have you tried? Were the norms finite or infinite? Oct 1 '20 at 18:38
• 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 $T$. Oct 1 '20 at 19:02
• 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. Jan 15 at 8:33

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

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

and

$$\norm{f}_\infty=\sup_{(0,\infty)}|f|$$

(if they exist)

$$p=1):$$ Consider the function

$$f(t)=\max\left(\frac{1}{\sqrt{t}}-1,0\right)$$

Then

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

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

$$\norm{f}_\infty=\infty$$

$$p=2):$$ Consider the function

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

Then

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

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

$$\norm{f}_\infty=\infty$$

$$p=\infty):$$ Consider the function

$$f(t)=\frac{1}{\sqrt{t+1}}$$

Then

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

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

$$\norm{f}_\infty=1$$

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

$$\frac{|f|^2}{M^2}=\left(\frac{|f|}{M}\right)^2<\frac{|f|}{M}$$

(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

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