Prove that l^p is a subset of l^q for all p,q from 1 to infinity

[cbarker1]

Homework Statement

Prove that l^p is a subset of l^q for all p,q from 1 to infinity. Then prove it is strict subset. First, prove that a^t<=a for all t,a in (0,1]. Then prove that finite sum of |x_i|^t<= the sum of |xi|.

Relevant Equations

a^t<=a for all a,t
p-norm's definition.

Dear everyone,

I am having trouble with this problem. I have convinced myself that the $a^t-a\leq 0$ is true. Now, I am trying to applying this inequality for the finite series and I don't know where to start. After that, proving that the p-norm is less or equal to the q-norm.

Thanks,
Cbarker1

 

[pasmith]

Is there a condition on p and q, such as q &lt; p? Otherwise you are being asked to prove l^p \subsetneq l^q \subsetneq l^p which is impossible.

If you have two sequences of non-negative numbers, with the property that each element of the first sequence is less than or equal to the corresponding element of the second sequence, what can you say about the sums of those sequences?

 

[jade_nick]

For this statement we will assume $p<q$. We can think of this as intuitively saying that only the small terms of a series matter for convergence granted the sequence it's built from converges to zero.

We can first address the strict inequality as it's easier. Consider $\{\left(\frac{1}{n}\right\)^{1/p}}_{n=1}^\infty$. We notice that applying the $p$-norm to this sequence is the $p$th root of the harmonic series, which as the harmonic series diverges, the norm diverges. Thus, this is not an element of $l^p$. On the otherhand we notice that $\sum_{i=1}^\infty n^{-\frac {q}{p}}$ converges by our $p$ test. This means we have an element of $l^q$ not in $l^p$.

Take $.4=.16^{.5}>.16$. This shows that your statement as formulated has a mistake, assuming you meant $t,a\in(0,1]$.

I'll assume you meant $a\leq1$ and $t\geq1$ as I believe this makes recovers the error. We recognize that $\gamma:=t-1\geq 0$. This allows us the say that as $a\leq 1$ we have that $a^{\gamma}\leq 1$ and $a^t=a\cdot a^{\gamma}\leq a\cdot 1=a$. This justifies our first statement.

Now suppose $b\in[0,1)$ and $p<q$ this means $1\leq \frac {q}{p}$ and $b^p\in[0,1)$. Applying our statement where $a=b^p$ and $t=\frac{q}{p}$ shows that $a^p=b>b^t=(a^p)^{p/q}=a^q$. We will use this iteration of our statement to prove that any element of $l^p$ is in $l^q$

Suppose we have a sequence $\{a_n\}\_{n=1}\in l^p$ then all but finitely many of our $l^p$ have absolute value less than $1$. If you have trouble justifying this recall that $\{a_n\}_{n=1}^\infty$ must converge to $0$ and use the $\d$ $\eps$ definition of convergence with $\eps=1/2$. Suppose all of our elements after $N-1$ have absolute value less than $1$. This means $\sum_{n=1}^{\infty}|a_{n}|^p=\sum_{i=1}^{N-1} |a_{n}|^p+\sum_{n=N}^\infty |a_{n}|^p$. We use the statement from our previous statement where $b=a_n$ and $t$ is as before on every element of the infinte sum to yield that $\sum_{n=1}^{\infty}|a_{n}|^p\geq \sum_{i=1}^{N-1} |a_{n}|^p+\sum_{n=N}^\infty |a_{n}|^q$. We have this sum is bounded by a constant. Adding the constant $\sum_{i=1}^{N-1} |a_{n}|^q-\sum_{i=1}^{N-1} |a_{n}|^p$ would still be bounded. Thus $\sum_{n=1}^\infty |a_n|^q$ is an increasing sequence bounded above by a constant value, meaning it's $q$th root is as well and $\{a_n\}_{n=1}^\infty$\in l^q.

You made one comment that I want to make a comment on. Unintuitively, even though every $l^p$ sequence is an $l^q$ sequence, quite often the p norm is smaller. The tail of the sequence is all that really matters for convergence, and the increased power just sends it to $0$ a little bit faster, but quite often almost all of the value of our sequence is comprise in the first few terms. Think about \{2^{1-n}\}_{n=1}^\infty. For our $p$ norm we get a value between $2^p$ and $2^p+1$ and the same thing for $2^q$ but $2^q$ is larger than $2^p+1$. Our $q$ norm shoots finitely many points to a much larger value, but there's only finitely many of such values. This is why we had to be careful in our proof to not say one is greater than the other, but instead talk of having an increasing sequence bounded by a constant.

 

[berkeman]

For this statement we will assume $p<q$. We can think of this as intuitively saying that only the small terms of a series matter for convergence granted the sequence it's built from converges to zero.

We can first address the strict inequality as it's easier. Consider $\{\left(\frac{1}{n}\right\)^{1/p}}_{n=1}^\infty$. We notice that applying the $p$-norm to this sequence is the $p$th root of the harmonic series, which as the harmonic series diverges, the norm diverges. Thus, this is not an element of $l^p$. On the otherhand we notice that $\sum_{i=1}^\infty n^{-\frac {q}{p}}$ converges by our $p$ test. This means we have an element of $l^q$ not in $l^p$.

<<snip>>

Welcome to PF. I sent you a DM just now with some hints on using LaTeX here at PF, so you will be able to fix up your post above.

Also, please keep in mind that we do not allow helpers to do the student's schoolwork for them. The student must do the bulk of the work. You can ask questions, give hints, find mistakes, etc. But in general you should not be doing the detailed work on a student's schoolwork here.

Having said that, since this thread you posted in is a couple years old, presumably the OP has finished that class and moved on, so posting a full solution is allowed in this case.

 

[FactChecker]

For this statement we will assume $p<q$.

IMO, the first duty of a person asking for help is to make a correct, accurate question. Part of the OP is trivially wrong if $p=q$.