Saturday 17 December 2016

A mind set of using linear algebra

Warning: Heavy use of LaTeX follows. Make sure you enabled Javascript to read those equations properly. It is highly recommended to enlarge the page so that the equations are not in congested form.

First of all, seasons greetings everyone. Let's see if I can get the customization article done before the end of 2016...


Readers probably know that I use linear algebra extremely heavily -- for casual blog posts, game studies, hardcore research and so on. It is really useful and it simplifies quite a lot of problems when you are stuck to prove it in a traditional way. You can find it in computational maths, topology, combinatorics, and now analysis.

Definition 1. Let $r\in \mathbb{N}_0$. Define $C^r$ to be the space of $r$-times differentiable functions, with the norm

$||f||_{C^r} = \sum _{|s|\leq r} ||D^sf||_{\infty}$

Definition 2. Let $r \in \mathbb{N}_0$ and $\alpha \in (0,1)$. Define $C^{r+\alpha}$ be the space so that the following equipped norm is finite:

$||f||_{C^{r+\alpha}} = \sum _{0\leq |s| < r} ||D^sf||_{\infty} + \sum _{|s|=r} H\ddot{o}l_{\alpha}(D^sf) < \infty $

Theorem 3. Some primitive results. There exist some constants $C>0$ so that:

1) If $f\in C^1$ then $||f||_{C^{\alpha}} \leq C ||f||_{\infty}^{1-\alpha} ||f||_{C^1}^{\alpha}$.

2) If $f\in C^{2+\alpha}$ then $||f||_{C^1} \leq C||f||_{C^{\alpha}}^{(\alpha +1)/2}||f||_{C^{2+\alpha}}^{(1-\alpha)/2}$.

In the rest of this article, $C$ represents some positive constants probably varying from line to line.

Proof. (1) is relatively easy to do using MVT:

$|f(x)-f(y)| = |f(x)-f(y)|^{1-\alpha} |f(x)-f(y)|^{\alpha}$

$\leq (2||f||_{\infty})^{1-\alpha}(||f||_{C^1}|x-y|)^{\alpha}$

$\leq C ||f||_{\infty}^{1-\alpha} ||f||_{C^1}^{\alpha}|x-y|^{\alpha}$

(2) involves some nasty analytical approximation so it will be skipped here.

We split the exponent $1$ into $(1-\alpha) + \alpha$, but what about harder interpolation estimates?

Theorem 4. If $f\in C^{2+\alpha}$ then $||f||_{C^1} \leq C ||f||_{\infty}^{\frac{1+\alpha}{2+\alpha}} ~ ||f||_{C^{2+\alpha}}^{\frac{1}{2+\alpha}}$ for some constant $C>0$.

Well, notice that these kinds of interpolation results share two common features: if $||f||_{C^p} \leq \prod ||f||_{C^{q_n}}^{r_n}$ then $\sum r_n =1 $ and $p = \sum q_nr_n$. These properties are preserved by the usual row operations!

Let $a,b,c,d$ be the exponent for the norm of $C^0, C^{\alpha}, C^1, C^{2+\alpha}$ respectively. Putting those 3 equations together we have the following matrix

$A = \begin{bmatrix} \alpha -1& 1 & -\alpha & 0 \\ 0 & -\frac{\alpha+1}{2} & 1 & \frac{\alpha-1}{2} \\ -\frac{1+\alpha}{2+\alpha} & 0 & 1 & -\frac{1}{2+\alpha} \end{bmatrix}$

which, upon reduction, has a zero row. Therefore theorem 4 is true.


If you want the real exponent to work with, use the linear dependency algorithm:

$A^t \sim \begin{bmatrix} 1& 0 & \frac{-\alpha -1}{\alpha ^2 + \alpha -2} \\ 0 & 1 & \frac{-2}{\alpha ^2 + \alpha -2} \\ 0 & 0 & 0\\0& 0& 0\end{bmatrix}$

The signs are not particularly useful because we have to care about the orientation of signs -- that flips if we take the negative power. Now upon such suitable exponent, we have the following from theorem 3:

$||f||_{C^{\alpha}}^{\frac{\alpha +1}{\alpha ^2+\alpha -2}} \geq C ||f||_{\infty}^{-\frac{1+\alpha}{2+\alpha}}~ ||f||_{C^1}^{\frac{\alpha (\alpha -1)}{\alpha ^2 + \alpha -2}}$

$||f||_{C^1}^{-\frac{2}{\alpha ^2+\alpha -2}} \leq C||f||_{C^{\alpha}}^{-\frac{\alpha +1}{\alpha ^2 + \alpha -2}}~||f||_{C^{2+\alpha}}^{\frac{1}{2+\alpha}}$

And, therefore, replicating the proof:

$||f||_{C^1} = (||f||_{C^1}^{-\frac{2}{\alpha ^2+\alpha -2}}~ )(||f||_{C^1}^{\frac{\alpha (\alpha -1)}{\alpha ^2 + \alpha -2}}~)$

$\leq C (||f||_{C^{\alpha}}^{-\frac{\alpha +1}{\alpha ^2 + \alpha -2}}~ ||f||_{C^{2+\alpha}}^{\frac{1}{2+\alpha}})(||f||_{C^{\alpha}}^{\frac{\alpha +1}{\alpha ^2+\alpha -2}}~ ||f||_{\infty}^{\frac{1+\alpha}{2+\alpha}})$

$ = C ||f||_{\infty}^{\frac{1+\alpha}{2+\alpha}} ~ ||f||_{C^{2+\alpha}}^{\frac{1}{2+\alpha}}$

You may actually find an easier solution to the above: just do substitution twice:

$||f||_{C^1} \leq C||f||_{C^{\alpha}}^{\frac{\alpha +1}{2}}~ ||f||_{C^{2+\alpha}}^{\frac{1-\alpha}{2}} \leq (||f||_{\infty}^{1-\alpha}~ ||f||_{C^1}^{\alpha})^{\frac{\alpha +1}{2}}~ ||f||_{C^{2+\alpha}}^{\frac{1-\alpha}{2}}$

which gives you the right answer after some rearrangement, but what about more complicated interpolation results? I simply leave two simple results here -- be warned that they are linearly independent from the above, so if they have rank 1 then you have to prove at least one of them using analytical results. The bad news is, they increase the rank of the system by 2. Can you still try to make use of linear algebra...?

Theorem 5. Let $f\in C^{2+\alpha}$, then the following holds for some $C>0$:

1): $||f||_{C^1} \leq C||f||_{\infty}^{\frac{\alpha}{\alpha +1}}~ ||f||_{1+\alpha}^{\frac{1}{\alpha +1}}$

2): $||f||_{C^2} \leq C||f||_{\infty}^{\frac{\alpha}{2+\alpha}}~ ||f||_{C^{2+\alpha}}^{\frac{2}{2+\alpha}}$

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