MHB Proving the Existence of Constants in a Summation Inequality

[evinda]

Hello! (Wave)

How can we show that there are constants $c_m$ such that:

$$\sum_{|a| \leq m} |\xi^a|^2 \leq (1+ |\xi|^2)^m \leq c_m \sum_{|a| \leq m} |\xi^a|^2$$

Could you give me a hint what we could do?

 

[I like Serena]

Hello! (Wave)

How can we show that there are constants $c_m$ such that:

$$\sum_{|a| \leq m} |\xi^a|^2 \leq (1+ |\xi|^2)^m \leq c_m \sum_{|a| \leq m} |\xi^a|^2$$

Could you give me a hint what we could do?

Hi evinda! (Smile)

What do we get if we expand $(1+ |\xi|^2)^m$ with the binomial expansion? (Wondering)

 

[evinda]

Hi evinda! (Smile)

What do we get if we expand $(1+ |\xi|^2)^m$ with the binomial expansion? (Wondering)

$(1+ |\xi|^2)^m= \sum_{k=0}^m \binom{m}{k} (|\xi|^k)^2$

What bound could we use for $\binom{m}{k}$ ?

 

[I like Serena]

$(1+ |\xi|^2)^m= \sum_{k=0}^m \binom{m}{k} (|\xi|^k)^2$

What bound could we use for $\binom{m}{k}$ ?

$$1 \le \binom{m}{k} \le \binom{m}{\lfloor m/2 \rfloor} \le m!$$
(Thinking)

 

[evinda]

$$1 \le \binom{m}{k} \le \binom{m}{\lfloor m/2 \rfloor} \le m!$$
(Thinking)

So we have $(1+ |\xi|^2)^m \leq m! \sum_{k=0}^m (|\xi|^k)^2$.

Is the latter equivalent to $m! \sum_{|\alpha| \leq m} |\xi^{\alpha}|^2$ even if $\alpha$ is a vector?

 

[I like Serena]

So we have $(1+ |\xi|^2)^m \leq m! \sum_{k=0}^m (|\xi|^k)^2$.

Is the latter equivalent to $m! \sum_{|\alpha| \leq m} |\xi^{\alpha}|^2$ even if $\alpha$ is a vector?

$\alpha$ shouldn't be a vector should it? (Wondering)
I'd expect $\alpha$ to be a number, and since it's used as index for a summation, I expect it to be an integer. (Thinking)

 

[evinda]

$\alpha$ shouldn't be a vector should it? (Wondering)
I'd expect $\alpha$ to be a number, and since it's used as index for a summation, I expect it to be an integer. (Thinking)

$\alpha$ is a multi-index, $\alpha=(\alpha_1, \dots, \alpha_n)$ and $|\alpha|=\alpha_1+ \dots+ \alpha_n$.

 

[Euge]

By the multinomial theorem,

$$(1 + |\xi|^2)^m = \sum_{|\beta| = m} \binom{m}{\beta}1^{\beta_0}(\xi_1^2)^{\beta_1}\cdots (\xi_n^2)^{\beta_n} = \sum_{\lvert \beta \rvert = m} \binom{m}{\beta} \lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2$$

where the sum is extended over all multi-indices $\beta\in \Bbb N_0^{n+1}$ with $\lvert \beta \rvert = m$. The index set above contains the multi-indices with first coordinate zero, which corresponds to the multi-indices $\alpha \in \Bbb N_0^n$ with $\lvert \alpha \rvert = n$. Hence, the above sum is greater than or equal to

$$\sum_{\lvert \alpha \rvert = m} \binom{m}{\alpha} \lvert\xi^\alpha \rvert^2 \ge \sum_{\lvert \alpha \rvert = m} \lvert \xi^\alpha\rvert^2$$

On the other hand, for each $(n+1)$-index $\beta = (\beta_0,\beta_1,\ldots, \beta_n)$, the $n$-index $\alpha = (\beta_1,\ldots, \beta_n)$ has $\lvert \alpha\rvert \le m$. Thus

$$\sum_{\lvert \beta \rvert = m} \binom{m}{\beta}\lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2 \le \sum_{\lvert \alpha \rvert \le m} \binom{m}{\alpha}|\xi^{\alpha}|^2 \le c_m \sum_{\lvert \alpha\rvert \le m} \lvert\xi^\alpha\rvert^2$$

where $c_m = \max\{\binom{m}{\alpha} : \alpha \in \Bbb N_0^n\}$.

We have now established the estimates

$$\sum_{\lvert \alpha\rvert \le m} \lvert \xi^\alpha\rvert^2 \le (1 + \lvert \xi\rvert^2)^m \le c_m \sum_{\lvert \alpha\rvert \le m} \lvert \xi^\alpha\rvert^2$$

as desired.

 

[evinda]

By the multinomial theorem,

$$(1 + |\xi|^2)^m = \sum_{|\beta| = m} \binom{m}{\beta}1^{\beta_0}(\xi_1^2)^{\beta_1}\cdots (\xi_n^2)^{\beta_n} = \sum_{\lvert \beta \rvert = m} \binom{m}{\beta} \lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2$$

Do we symbolize $1^{\beta_0}(\xi_1^2)^{\beta_1}\cdots (\xi_n^2)^{\beta_n}$ by $\lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2$ ?

where the sum is extended over all multi-indices $\beta\in \Bbb N_0^{n+1}$ with $\lvert \beta \rvert = m$. The index set above contains the multi-indices with first coordinate zero, which corresponds to the multi-indices $\alpha \in \Bbb N_0^n$ with $\lvert \alpha \rvert = n$.

Why does it hold that $\lvert \alpha \rvert = n$ and not $\lvert \alpha \rvert = m- \lvert \beta_0 \rvert$ ?

On the other hand, for each $(n+1)$-index $\beta = (\beta_0,\beta_1,\ldots, \beta_n)$, the $n$-index $\alpha = (\beta_1,\ldots, \beta_n)$ has $\lvert \alpha\rvert \le m$. Thus

$$\sum_{\lvert \beta \rvert = m} \binom{m}{\beta}\lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2 \le \sum_{\lvert \alpha \rvert \le m} \binom{m}{\alpha}|\xi^{\alpha}|^2 \le c_m \sum_{\lvert \alpha\rvert \le m} \lvert\xi^\alpha\rvert^2$$

where $c_m = \max\{\binom{m}{\alpha} : \alpha \in \Bbb N_0^n\}$.

Why do we deduce from this that $\sum_{\lvert \beta \rvert = m} \binom{m}{\beta}\lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2 \le \sum_{\lvert \alpha \rvert \le m} \binom{m}{\alpha}|\xi^{\alpha}|^2 $ and not that $\sum_{\lvert \beta \rvert = m} \binom{m}{\beta}\lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2 \geq \sum_{\lvert \alpha \rvert \le m} \binom{m}{\alpha}|\xi^{\alpha}|^2 $ ?

 

[Euge]

Do we symbolize $1^{\beta_0}(\xi_1^2)^{\beta_1}\cdots (\xi_n^2)^{\beta_n}$ by $\lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2$ ?

No. It follows from the calculation

$$1^{\beta_0}(\xi_1^2)^{\beta_1}\cdots (\xi_n^2)^{\beta_n} = (\xi_1^{\beta_1})^2\cdots (\xi_n^{\beta_n})^2 = \lvert\xi_1^{\beta_1}\cdots \xi_n^{\beta_n}\rvert^2 = \lvert \xi^{(\beta_1,\dots,\beta_n)}\rvert^2$$

Why does it hold that $\lvert \alpha \rvert = n$ and not $\lvert \alpha \rvert = m- \lvert \beta_0 \rvert$ ?

There was a typo there -- it's supposed to be $\lvert \alpha \rvert = m$. Don't forget that we considered all $(n+1)$-indices whose first coordinate is zero, i.e., $\beta_0 = 0$. Also, since $\beta_0\in \Bbb N_0$, it is unnecessary to write $\lvert \beta_0\rvert$ - just write $\beta_0$.

Why do we deduce from this that $\sum_{\lvert \beta \rvert = m} \binom{m}{\beta}\lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2 \le \sum_{\lvert \alpha \rvert \le m} \binom{m}{\alpha}|\xi^{\alpha}|^2 $ and not that $\sum_{\lvert \beta \rvert = m} \binom{m}{\beta}\lvert \xi^{(\beta_1,\ldots, \beta_n)}\rvert^2 \geq \sum_{\lvert \alpha \rvert \le m} \binom{m}{\alpha}|\xi^{\alpha}|^2 $ ?

Recall the follows property of summation: If $B$ is a subset of $A$, then

$$\sum_{\alpha\in B} x_\alpha \le \sum_{\alpha\in A} x_{\alpha}$$

The set $A$ of indices $\alpha = (\beta_1,\ldots, \beta_n)$ with $\lvert \alpha\rvert \le m$ contains the set $B$ of indices $\beta = (\beta_0,\beta_1,\ldots,\beta_n)$ with $\lvert \alpha\rvert = m - \beta_0,$ i.e., $\beta_0 + \beta_1 + \cdots + \beta_n = m$, or $\lvert \beta\rvert = m$. So the inequality I wrote follows.

 

[evinda]

No. It follows from the calculation

$$1^{\beta_0}(\xi_1^2)^{\beta_1}\cdots (\xi_n^2)^{\beta_n} = (\xi_1^{\beta_1})^2\cdots (\xi_n^{\beta_n})^2 = \lvert\xi_1^{\beta_1}\cdots \xi_n^{\beta_n}\rvert^2 = \lvert \xi^{(\beta_1,\dots,\beta_n)}\rvert^2$$

There was a typo there -- it's supposed to be $\lvert \alpha \rvert = m$. Don't forget that we considered all $(n+1)$-indices whose first coordinate is zero, i.e., $\beta_0 = 0$. Also, since $\beta_0\in \Bbb N_0$, it is unnecessary to write $\lvert \beta_0\rvert$ - just write $\beta_0$.

A ok (Nod)

Recall the follows property of summation: If $B$ is a subset of $A$, then

$$\sum_{\alpha\in B} x_\alpha \le \sum_{\alpha\in A} x_{\alpha}$$

The set $A$ of indices $\alpha = (\beta_1,\ldots, \beta_n)$ with $\lvert \alpha\rvert \le m$ contains the set $B$ of indices $\beta = (\beta_0,\beta_1,\ldots,\beta_n)$ with $\lvert \alpha\rvert = m - \beta_0,$ i.e., $\beta_0 + \beta_1 + \cdots + \beta_n = m$, or $\lvert \beta\rvert = m$. So the inequality I wrote follows.

How can the set $A$ of indices $\alpha = (\beta_1,\ldots, \beta_n)$ with $\lvert \alpha\rvert \le m$ contain the set $B$ of indices $\beta = (\beta_0,\beta_1,\ldots,\beta_n)$ although the indices of the first set contain $n$ components but the second $n+1$ ?

 

[Euge]

How can the set $A$ of indices $\alpha = (\beta_1,\ldots, \beta_n)$ with $\lvert \alpha\rvert \le m$ contain the set $B$ of indices $\beta = (\beta_0,\beta_1,\ldots,\beta_n)$ although the indices of the first set contain $n$ components but the second $n+1$ ?

Let me be more precise, and this time I'll write out the summations so you can keep track.

$$\sum_{\lvert \beta\rvert = m} \binom{m}{\beta} \lvert \xi^{(\beta_1,\ldots,\beta_n)}\rvert^2 =
\sum_{\beta_0 \le m} \sum_{\lvert (\beta_1,\ldots,\beta_n)\rvert = m - \beta_0} \binom{m}{\beta} \lvert \xi^{(\beta_1,\ldots,\beta_n)}\rvert^2 \frac{1}{\beta_0!}\binom{m}{(\beta_1,\ldots,\beta_n)}\lvert \xi^{(\beta_1,\ldots,\beta_n)}\rvert^2 \le \sum_{\beta_0\le m} \sum_{\lvert \alpha\rvert = m - \beta_0} \binom{m}{\alpha} \lvert \xi^\alpha\rvert^2 = \sum_{\lvert \alpha\rvert \le m} \binom{m}{\alpha} \lvert \xi^{\alpha}\rvert^2$$

 

[evinda]

Let me be more precise, and this time I'll write out the summations so you can keep track.

$$\sum_{\lvert \beta\rvert = m} \binom{m}{\beta} \lvert \xi^{(\beta_1,\ldots,\beta_n)}\rvert^2 =
\sum_{\beta_0 \le m} \sum_{\lvert (\beta_1,\ldots,\beta_n)\rvert = m - \beta_0} \binom{m}{\beta} \lvert \xi^{(\beta_1,\ldots,\beta_n)}\rvert^2 \frac{1}{\beta_0!}\binom{m}{(\beta_1,\ldots,\beta_n)}\lvert \xi^{(\beta_1,\ldots,\beta_n)}\rvert^2 \le \sum_{\beta_0\le m} \sum_{\lvert \alpha\rvert = m - \beta_0} \binom{m}{\alpha} \lvert \xi^\alpha\rvert^2 = \sum_{\lvert \alpha\rvert \le m} \binom{m}{\alpha} \lvert \xi^{\alpha}\rvert^2$$

How did you get this: $ \frac{1}{\beta_0!}\binom{m}{(\beta_1,\ldots,\beta_n)}\lvert \xi^{(\beta_1,\ldots,\beta_n)}\rvert^2$ at the sum?

 

[Euge]

How did you get this: $ \frac{1}{\beta_0!}\binom{m}{(\beta_1,\ldots,\beta_n)}\lvert \xi^{(\beta_1,\ldots,\beta_n)}\rvert^2$ at the sum?

It comes from the identity

$$\binom{m}{\beta} = \frac{1}{\beta_0!} \binom{m}{(\beta_1,\ldots,\beta_n)}$$

To explain the notation on the right, I'm using $\binom{m}{(\beta_1,\ldots,\beta_n)}$ as a shortand for $\frac{m!}{\beta!}$ (remember $\beta! = (\beta_1)!\cdots (\beta_n)!$). This is an extension of the usual meaning of the multinomial coefficient, where it is required that $\beta_1 + \cdots + \beta_n = m$. Sorry for the confusion.

Using the normal notation, the inequality we speak of is

$$\sum_{\lvert \beta\rvert = m} \binom{m}{\beta}\lvert \xi^\beta\rvert^2 \le \sum_{\lvert \alpha\rvert \le m} \frac{m!}{\alpha!}\lvert \xi^\alpha\rvert^2$$

 

[evinda]

It comes from the identity

$$\binom{m}{\beta} = \frac{1}{\beta_0!} \binom{m}{(\beta_1,\ldots,\beta_n)}$$

To explain the notation on the right, I'm using $\binom{m}{(\beta_1,\ldots,\beta_n)}$ as a shortand for $\frac{m!}{\beta!}$ (remember $\beta! = (\beta_1)!\cdots (\beta_n)!$). This is an extension of the usual meaning of the multinomial coefficient, where it is required that $\beta_1 + \cdots + \beta_n = m$. Sorry for the confusion.

Using the normal notation, the inequality we speak of is

$$\sum_{\lvert \beta\rvert = m} \binom{m}{\beta}\lvert \xi^\beta\rvert^2 \le \sum_{\lvert \alpha\rvert \le m} \frac{m!}{\alpha!}\lvert \xi^\alpha\rvert^2$$

I got it... Thank you very much! (Smile)