Analysis Tricks Part 1

by | Dec 28, 2018 | Problem Solving, Pure Maths | 0 comments

I recently learnt a really neat trick when it comes to proving two quantities are equivalent. It started of with my supervisor claiming that for some \xi \in \mathbb{R}^n and m\in \mathbb{R},

(1)   \begin{equation*} \Sigma_{\abs{\alpha}\leq m} \abs{\xi^{\alpha}}^2 \simeq 1+\Sigma_{i=1}^{n}\xi_i^{2m}, \end{equation*}

is obvious. To me, it was not obvious at all, and may not be so obvious to you either. But what my supervisor was referring to was a trick that saves a lot of time when it comes to analysis, and so today, I will be talking about it.

I will first prove 1 before I show you the trick. We firstly need to understand the notation in 1;

    \[\alpha=(\alpha_1,\dots,\alpha_n)~~~\alpha_i>0\]


is called a \textit{multi index} and it is used to raise vectors to powers. For some x \in \mathbb{R}^n we define

    \[x^{\alpha}=x_1^{\alpha_1}\cdot\dots\cdot x_n^{\alpha_n}=\Pi_{i=1}^{n}x_i^{\alpha_i}.\]

The degree of \alpha is \abs{\alpha}=\Sigma_{i=1}^{n}\alpha_i

With this in hand, we can re-write 1 as

(2)   \begin{equation*} \Sigma_{\abs{\alpha}\leq m}\abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}} \simeq 1+\Sigma_{i=1}^{n}\xi_i^{2m}. \end{equation*}

Now, for the equivalence symbol; x\simeq y means we can find two constants A and B such that

    \[Ay\leq x\leq By,\]


and we say that x is equivalent to y.

We shall first prove that there exists a constant C such that

    \[\Sigma_{\abs{\alpha}\leq m}\abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}} \geq C(1+\Sigma_{i=1}^{n}\xi_i^{2m}).\]

To do this, we take a small subset of the multi-indices that satisfy \abs{\alpha}\leq m, the index \alpha_0 which has zero in every entry and also the indices \alpha_i for which

(3)   \begin{equation*} (\alpha_i)_j= \begin{cases} m ~ \text{if}~i=j\ 0 ~\text{if}~i\neq j \end{cases}. \end{equation*}


We call the set of these indices K and we note that

(4)   \begin{align*} \Sigma_{\abs{\alpha}\leq m}\abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}} &\geq \Sigma_{\alpha\in K}\abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}}\\ &=1+\Sigma_{i=1}^{n}\xi_i^{2m}. \end{align*}

So taking C=1, we have proved the first direction. Now, we wish to prove that there exists a constant D such that

    \[\Sigma_{\abs{\alpha}\leq m}\abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}} \leq D(1+\Sigma_{i=1}^{n}\xi_i^{2m}).\]

Prepare yourself for trick number 1 that will make your life so much easier, because the left hand side is just a finite sum, we only need to control the term \abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}} by (1+\Sigma_{i=1}^{n}\xi_i^{2m}). Then we can take D=N where N is the number of elements in the set {\alpha: \abs{\alpha}\leq m} and the result will follow.

We will assume without loss of generality, that \abs{\xi_i} is an increasing sequence. We begin by applying the Cauchy-Schwartz inequality:

(5)   \begin{align*} \abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}} &\leq \Pi_{i=1}^{n} \abs{\xi_{i}}^{2\alpha_i}\\ &\leq \Pi_{i=1}^{n} \abs{\xi_{n}}^{2\alpha_i}\\ &=\abs{\xi_{n}}^{2\Sigma_{i=1}^n\alpha_i}. \end{align*}

We have two cases, if \abs{\xi_n}<1 then we have that

(6)   \begin{equation*} \abs{\xi_{n}}^{2\Sigma_{i=1}^n\alpha_i}<1 \end{equation*}

and the result follows for this case.

However, if \abs{\xi_n}\geq1 then knowing that 2\Sigma_{i=1}^n\alpha_i\leq2m we can write

(7)   \begin{equation*} \abs{\xi_{n}}^{2\Sigma_{i=1}^n\alpha_i}\leq\abs{\xi_{n}}^{2m}\leq\Sigma_{i=1}^n\xi_i^{2m}, \end{equation*}

and the result follows.

So we have proved that

(8)   \begin{equation*} \Sigma_{\abs{\alpha}\leq m}\abs{\xi^{\alpha}}^2 \simeq 1+\Sigma_{i=1}^{n}\xi_i^{2m}, \end{equation*}


but there is an intuitive way to look at this problem.

Firstly, consider the finite sum for x,a >0

(9)   \begin{equation*} x^a+x^{a+1}+\dots+x^{a+n}. \end{equation*}

If x<1, we can control the sum by nx^a but if x\geq1 we can control it by nx^{a+n}, and so in general,

(10)   \begin{equation*} x^a+x^{a+1}+\dots+x^{a+n}\simeq x^a+x^{a+n}. \end{equation*}

What we did was look at extreme cases and saw how all the other cases in between them could be controlled, if we controlled the extreme ones. This is the trick! Now let’s apply it to 2 which is

    \[\Sigma_{\abs{\alpha}\leq m}\abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}} \simeq 1+\Sigma_{i=1}^{n}\xi_i^{2m}.\]

Again, we note that we only need to control the term that is being summed, and not the entire sum, as we saw from the proof. We now look at the extreme cases, if \alpha is the index with zero in every entry, then

    \[\abs{\Pi_{i=1}^{n} \xi_{i}^{2\alpha_i}}=1,\]

and if \abs{\alpha}=m we can control these cases by

    \[\Sigma_{i=1}^{n}\xi_i^{2m},\]

since \xi_i^{2m}\leq\Sigma_{i=1}^{n}\xi_i^{2m} for all i. Hence, by an intuitive path, we have arrived at 2.

We note that in this trick, we only care about the order of these expressions, we look at the extreme cases of the order and note that we can control everything in between the extreme cases.

Stay tuned to the next blog post to see this trick, (and many more!!) in action.

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