# Are the p-norm and and the p'-norm equivalent?

1. Jun 2, 2014

### dirk_mec1

1. The problem statement, all variables and given/known data
Are the p-norm and and the p'-norm equivalent?

With
$$1 \leq p < \infty$$ and $$1 \leq p' < \infty$$

2. Relevant equations

Two norms p and q are equivalent if there is a M>0 and m>0 such that $$||x||_p \leq M ||x||_q$$ and

$$||x||_q \leq m ||x||_p$$

3. The attempt at a solution
I do not seem to get started. How can I find the constants?

2. Jun 2, 2014

### LCKurtz

1. Tell us what vector space $x$ is in.
2. Show us what you have tried.

3. Jun 2, 2014

### dirk_mec1

$$x \in \mathbb{R} ^k$$

Suppose I take k = 2 and I presume p>q (I do not like the accent so I choose to use q) then I need to prove that:

$$(|x_1|^p + |x_2|^p)^{1/p} \leq M (|x_1|^q + |x_2|^q)^{1/q}$$

This is not correct but if I assume the vector elements to be larger than one then:

I know that $|x_i|^p \geq |x_i|^q$ with i=1,2.

I also know that $|x_i|^{1/p} \leq |x_i|^{1/q}$ with i=1,2.

Then I'm stuck.

Last edited: Jun 2, 2014
4. Jun 2, 2014

### jbunniii

Suppose $1 \leq p < q < \infty$.

If $\|x\|_p = 1$, can you find an upper bound for $|x_i|^p$? What does this imply about $\|x\|_q$ in this case?

5. Jun 3, 2014

### dirk_mec1

if $||x||_p =1$ then $|x_i|^p \leq 1$ which implies then that $|x_i|^q \leq |x_i|^p$ for i = 1,...,k but then:

$$|x_1|^q +... + |x_k|^q \leq |x_1|^p +... + |x_k|^p$$

taking root at both sides

$$(|x_1|^q +... + |x_k|^q)^{1/p} \leq (|x_1|^p +... + |x_k|^p)^{1/p}$$

but the RHS is 1

$$(|x_1|^q +... + |x_k|^q)^{1/p} \leq 1$$

taking root at both sides

$$||x||_q^{1/p} \leq 1$$

This has an lower bound

$$||x||_q \leq ||x||_q^{1/p}$$

So the conclusion is that:

$||x||_q \leq ||x||_p$ with $||x||_p=1$

Suppose the vector x has length A (in the p-norm) then we can adapt this to:

$||x||_q \leq ||x||_p \cdot A^{1/q}$

Is this correct?

Last edited: Jun 3, 2014
6. Jun 3, 2014

### jbunniii

OK so far.

OK, here you're using the fact that if you take the right hand side to the $p$'th power, then the result is smaller since $\|x\|_q^{1/p} \leq 1$.

You can save a few steps if you avoid taking $p$'th roots. If $\|x\|_p = 1$, then $|x_1|^p +... + |x_k|^p = 1$, so
$$|x_1|^q +... + |x_k|^q \leq |x_1|^p +... + |x_k|^p = 1$$
and therefore
$$\|x\|_q = (|x_1|^q +... + |x_k|^q)^{1/q} \leq 1 = \|x\|_p$$

No. If $\|x\|_p = A$, then we can write $x = Au$ where $\|u\|_p = 1$. Then:
$$\|x\|_q = \|Au\|_q = A\|u\|_q \leq ???$$
Now apply the result obtained above.

For the other inequality $||x||_p \leq M ||x||_q$, I recommend that you once again start with the case $\|x\|_p = 1$. In that case, you need to prove that there is some $M$ for which
$$\|x\|_q \geq \frac{1}{M}$$
The key idea here is that if $\|x\|_p = 1$, it's not possible for ALL of the elements of $x$ to be small. Can you find a lower bound for $\max |x_i|$?

7. Jun 5, 2014

### dirk_mec1

$$\|x\|_q = \|Au\|_q = A\|u\|_q \leq A\|u\|_p = \|Au\|_p= ||x||_p$$

So m=1 in this case?

I think that the lower bound for $\mbox{max}|x_i| = (\frac{1}{k})^{1/p}$

But I don't understand how to proceed.

Last edited: Jun 5, 2014
8. Jun 5, 2014

### jbunniii

Yes.

Yes, this is correct. Sketch of proof: if $|x_i| < (1/k)^{1/p}$ for all $i$, then $|x_i|^p < 1/k$ for all $i$, so $|x_1|^p + \cdots + |x_k|^p < 1$, contradicting the fact that $\|x\|_p = 1$. So for at least one $i$ we must have $|x_i| > (1/k)^{1/p}$.

For the next step, note that the above implies (for the same $i$) that $|x_i|^q > (1/k)^{q/p}$. What does this imply about $\|x\|_q$?

9. Jun 10, 2014

### dirk_mec1

I think it implies that $||x||_q \geq \left(\frac{1}{k} \right)^{q/p}$

Now suppose $||x||_p=A$

We can write $x=Au$ with $||u||_p= 1$ then we get

$||x||_q =||Au||_q =A ||u||_q \geq A \left(\frac{1}{k} \right)^{q/p} = M\cdot A||u||_p = M\cdot ||Au||_p = M ||x||_p$ with $M= \left(\frac{1}{k} \right)^{q/p}$

Is this correct?

Last edited: Jun 10, 2014
10. Jun 10, 2014

### jbunniii

Close but not quite right:
$$\|x\|_q^q = |x_1|^q + \cdots + |x_k|^q \geq |x_i|^q > (1/k)^{q/p}$$
where $i$ is as in my previous post. Taking $q$'th roots of both sides, we get
$$\|x\|_q > (1/k)^{1/p}$$
The rest of your post is fine if you make the corresponding adjustment.

11. Jun 10, 2014

### dirk_mec1

Ok got it thanks.