Proof of this recursion relation

Question

Let $f_J: [0,1] \rightarrow \mathbb{R}$, for $J = 1,2,3,\ldots$. That converges to a function, $f$. For J = 1, define $f_1(t) = t$. For $J > 1$, define a piecewise linear function, $$f_{J+1} = \begin{cases} f_J(t) \quad \forall t \in X_{J} \\ f_J(t) + (-1)^{k+J}2^{\frac{-1}{2}(J+2)} \quad \forall t = (2k-1)2^{-(J+1)} \quad k = 1, \ldots, 2^{J} \end{cases}$$ where $X_{J} =\{0,2^{-J},2\times 2^{-J}, 3\times 2^{-J}, \ldots , 1\}$.

Consider a point $t = k2^{-J}$ for some $J > 0$ and $0 \leq k < 2^{J}$. Show that for any such point, $$|f_{J'}(t + 2^{-J'}) - f_{J'}(t)| \geq \frac{2^{-J'/2}}{8}$$ for some $J' \in \{J+1, J+2\}$

Isn't it missing a part in the definition of $f_J$, it's only defined in discrete points, what about the rest of [0,1] ? — 2017-02-13
I apologize for the mistake, it should be a linear function connecting the end points. So connecting the discrete points with straight lines — 2017-02-13
With the definitions you present, both $f_{J+1}(t+2^{-(J+1)}$ and $f_{J+2}(t+2^{-(J+2)}$ do not have defined values. For example, $$f_{J+1}(k2^{-J}+2^{-(J+1)}) = f_J((2k+1)2^{2^{-(J+1)}+\mbox { other term}$$ and $f_J((2k+1)2^{-(J+1)}$ is not defined anywhere other than for $J=1$. — 2017-02-13
@MarkFischler It should be defined because it is simply a straight line connecting the discrete points, so the new point should simply be a midpoint of the two end points — 2017-02-13

user id:399263 13k · Accepted Answer

I think we'll ease the problem introducing a few notations.

$X_j=\{0,\frac{1}{2^j},\frac{2}{2^j},\frac{3}{2^j},...,1\}$

$Y_{j+1}=\{\frac{1}{2^{j+1}},\frac{3}{2^{j+1}},\frac{5}{2^{j+1}},...,\frac{2^{j+1}-1}{2^{j+1}}\}=X_{j+1}-X_j$

Let's define points in $X_j$ : $X_j(k)=\frac{k}{2^j}$ for $k=0\,..\,2^j$

Let's define points in $Y_j$ : $Y_{j+1}(k)=\frac{2k+1}{2^{j+1}}$ for $k=0\,..\,2^j-1$. (easier with 2k+1 than 2k-1).

First let's notice that $Y_{j+1}$ points are interlaced with $X_j$ points:

$X_j(k)

So $t\in Y_{j+1}\Rightarrow f_j(t)=f_j(Y_{j+1}(k))=\frac{f_j(X_j(k))+f_j(X_j(k+1))}{2}\quad$ since $f_j$ is linear in this interval.

And also since both $X_j$ and $Y_{j+1}$ are subsets of $X_{j+1}$ we have $\begin{cases} X_{j}(k)=X_{j+1}(2k)\\ Y_{j+1}(k)=X_{j+1}(2k+1)\\ \end{cases}$

We now have a more precise definition of $f_{j+1}$ :

$\left\{\begin{array}{lll} t\in X_j & t=X_j(k),\ k=0\,..\,2^j & f_{j+1}(t)=f_j(X_j(k)) \\ t\in Y_{j+1} & t=Y_{j+1}(k),\ k=0\,..\,2^j-1 & f_{j+1}(t)=\frac{f_j(X_j(k))+f_j(X_j(k+1))}{2}+\frac{(-1)^{k+1+j}}{2^\frac{j+2}{2}} \\ \end{array}\right.$

Now considering your question regarding $A(J)=f_J(t+2^{-J})-f_J(t)$ for $t=2^{-j}k$.

$t=2^{-j}k=X_j(k)$ for $k\neq2^j$ or $t\neq1$

For $J=j+1$ we have $t+2^{-J}=\frac{k}{2^j}+\frac{1}{2^{j+1}}=\frac{2k+1}{2^{j+1}}=Y_{j+1}(k)$

For $J=j+2$ we have $t+2^{-J}=\frac{k}{2^j}+\frac{1}{2^{j+2}}=\frac{2(2k)+1}{2^{j+2}}=Y_{j+2}(2k)$

$A(j+1)=f_{j+1}(Y_{j+1}(k))-f_{j+1}(X_j(k))=\frac{f_j(X_j(k+1))-f_j(X_j(k))}{2}+\frac{(-1)^{k+1+j}}{2^\frac{j+2}{2}}$

$A(j+2)=f_{j+2}(Y_{j+2}(2k))-f_{j+2}(X_j(k))=\frac{f_{j+1}(X_{j+1}(2k+1))+f_{j+1}(X_{j+1}(2k))}{2}+\frac{(-1)^{k+2+j}}{2^\frac{j+3}{2}}-f_j(X_j(k))=\frac{f_{j+1}(Y_{j+1}(k))+f_j(X_j(k))}{2}+...=\frac{f_j(X_j(k+1))-f_j(X_j(k))}{4}+\frac12\frac{(-1)^{k+1+j}}{2^\frac{j+2}{2}}+\frac{(-1)^{k+2+j}}{2^\frac{j+3}{2}}$

So we see that what's interest us is evaluating the difference of the values of $f_j$ taken in two consecutive points of $X_j$.

We have also that $2A(j+2)-A(j+1)=\frac{(-1)^{k+j}}{2^\frac{j+1}{2}}$ for $j\ge0$.

Let's call $a_n=(\sqrt 2)^nA(n)\ $ for $n\ge 1$

$(\sqrt 2)^{n+1}(2A(n+2)-A(n+1))=\sqrt 2\,a_{n+2}-a_{n+1}=(-1)^{k+n}$ $=-(-1)^{k+n-1}=-(\sqrt 2\,a_{n+1}-a_{n})$

And we get the double recurrence relation : $(\sqrt 2)a_{n+2}+(\sqrt 2-1)a_{n+1}+a_n=0$.

But $\Delta=(\sqrt 2-1)^2-4\sqrt 2=-(6\sqrt 2-3)<0$ is not so nice...

$|A(J)|\ge \frac{2^{-J/2}}{8}\iff |8a_J|\ge1$ for some $J$.

I'm stuck there too. It's a pity because this idea seemed promising, theoretically $a_n$ is completely determined by the recurrence, but practically the bad $\Delta$ ruins every chance to extract some practical stuff about $a_n$.

It seems, we are good for actually finding an explicit expression for $f_j(X_j(k))$.

To be continued...

Here is the function $f_7$, it looks like a nice fractal bumpy thing.

I was tired of making no progress, so since the values are in $\mathbb Q[\sqrt 2]=\{\frac{a+b\sqrt 2}{c}\ |\ (a,b,c)\in\mathbb Z^3\}$, I programmed arithmetic operations in this field, and let my computer calculate the values of $f_j$. Please find below a link to download the results.

Computed values of $f_j(X_j(k))$ : text zipped file

 X1
 |DIFF_MIN| =  5.000000e-01   1/2
 |DIFF_MAX| =  5.000000e-01   1/2
 |DIFF_MIN| / 2^(-j/2)/8 =  5.656854e+00
 |DIFF_MAX| / 2^(-j/2)/8 =  5.656854e+00
 X2
 |DIFF_MIN| =  1.035534e-01   (1-v2)/4
 |DIFF_MAX| =  6.035534e-01   (1+v2)/4
 |DIFF_MIN| / 2^(-j/2)/8 =  1.656854e+00
 |DIFF_MAX| / 2^(-j/2)/8 =  9.656854e+00
 X3
 |DIFF_MIN| =  5.177670e-02   (-1+v2)/8
 |DIFF_MAX| =  5.517767e-01   (3+v2)/8
 |DIFF_MIN| / 2^(-j/2)/8 =  1.171573e+00
 |DIFF_MAX| / 2^(-j/2)/8 =  1.248528e+01
 X4
 |DIFF_MIN| =  2.588835e-02   (-1+v2)/16
 |DIFF_MAX| =  4.526650e-01   (3+3v2)/16
 |DIFF_MIN| / 2^(-j/2)/8 =  8.284271e-01
 |DIFF_MAX| / 2^(-j/2)/8 =  1.448528e+01
 X5
 |DIFF_MIN| =  1.294417e-02   (-1+v2)/32
 |DIFF_MAX| =  3.513325e-01   (7+3v2)/32
 |DIFF_MIN| / 2^(-j/2)/8 =  5.857864e-01
 |DIFF_MAX| / 2^(-j/2)/8 =  1.589949e+01
 X6
 |DIFF_MIN| =  1.110435e-03   (7-5v2)/64
 |DIFF_MAX| =  2.640546e-01   (7+7v2)/64
 |DIFF_MIN| / 2^(-j/2)/8 =  7.106781e-02
 |DIFF_MAX| / 2^(-j/2)/8 =  1.689949e+01
 X7
 |DIFF_MIN| =  5.552173e-04   (7-5v2)/128
 |DIFF_MAX| =  1.945273e-01   (15+7v2)/128
 |DIFF_MIN| / 2^(-j/2)/8 =  5.025253e-02
 |DIFF_MAX| / 2^(-j/2)/8 =  1.760660e+01
 X8
 |DIFF_MIN| =  2.776086e-04   (7-5v2)/256
 |DIFF_MAX| =  1.414578e-01   (15+15v2)/256
 |DIFF_MIN| / 2^(-j/2)/8 =  3.553391e-02
 |DIFF_MAX| / 2^(-j/2)/8 =  1.810660e+01
 X9
 |DIFF_MIN| =  1.388043e-04   (7-5v2)/512
 |DIFF_MAX| =  1.019789e-01   (31+15v2)/512
 |DIFF_MIN| / 2^(-j/2)/8 =  2.512627e-02
 |DIFF_MAX| / 2^(-j/2)/8 =  1.846016e+01
 X10
 |DIFF_MIN| =  6.940216e-05   (7-5v2)/1024
 |DIFF_MAX| =  7.308654e-02   (31+31v2)/1024
 |DIFF_MIN| / 2^(-j/2)/8 =  1.776695e-02
 |DIFF_MAX| / 2^(-j/2)/8 =  1.871016e+01
 X11
 |DIFF_MIN| =  5.953764e-06   (-41+29v2)/2048
 |DIFF_MAX| =  5.216827e-02   (63+31v2)/2048
 |DIFF_MIN| / 2^(-j/2)/8 =  2.155493e-03
 |DIFF_MAX| / 2^(-j/2)/8 =  1.888693e+01
 X12
 |DIFF_MIN| =  2.976882e-06   (-41+29v2)/4096
 |DIFF_MAX| =  3.713268e-02   (63+63v2)/4096
 |DIFF_MIN| / 2^(-j/2)/8 =  1.524164e-03
 |DIFF_MAX| / 2^(-j/2)/8 =  1.901193e+01
 X13
 |DIFF_MIN| =  1.488441e-06   (-41+29v2)/8192
 |DIFF_MAX| =  2.637884e-02   (127+63v2)/8192
 |DIFF_MIN| / 2^(-j/2)/8 =  1.077746e-03
 |DIFF_MAX| / 2^(-j/2)/8 =  1.910032e+01
 X14
 |DIFF_MIN| =  7.442205e-07   (-41+29v2)/16384
 |DIFF_MAX| =  1.871369e-02   (127+127v2)/16384
 |DIFF_MIN| / 2^(-j/2)/8 =  7.620818e-04
 |DIFF_MAX| / 2^(-j/2)/8 =  1.916282e+01

$DIFF=f_j(X_j(k+1)-f_j(X_j(k))\simeq A(j)$

As you can notice DIFF_MAX seems to behave nicely and the ratio with $\frac{2^-(j/2)}{8}$ seems to converge to some fixed positive value.

Yet your question was about DIFF_MIN, and this time it is not ok at all, the ratio clearly decrease to $0$ and faster than expected.

Rem: this cannot seem to be fixable easily, because even if I take the ratio with $2^{-j}/8$ instead of $2^{-j/2}/8$ this is still decreasing to $0$ (not showed here, but I've made the test for myself).

The good ratio appears to be $\bbox[5px,border:2px solid red]{2^{-3(j+1)/2}}$ It holds till $j=25$, but I cannot test it beyond because of memory limitations.

So I think there is a mistake in your minoration statement!

Proof of this recursion relation

1 Answers 1

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