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$$ \begin{pmatrix} h(x-a) \\ h(x)\\ h(x+a) \end{pmatrix} = G ^{-1}(x) \begin{pmatrix} 0 \\ b\\ 0 \end{pmatrix} ,\quad x\in\biggl[-\frac{a}{2},\frac{a}{2}\biggr], $$

Remark 2.5

We note that this approach is tailored to the region of parameters \((a,b)\) in Theorem Outlet Discounts Buy Cheap Ebay Missguided Ruched Detail Bardot Mini knitted Dress gSYyGohk
. For example, it does not apply to the region considered in Proposition Stella McCartney White Georgie Donket Print Top Unisex Size 1518 Sale Sneakernews pbUBPU
(ii). In fact, if \(0< b\leq\frac{2}{N+a}\) , then the first row of \(G(x)\) for \(x\in [\frac{N}{2}-\frac{1}{b}+a,\frac{a}{2}]\) is the zero vector, and the third row of \(G(x)\) for \(x\in [-\frac{a}{2},\frac{1}{b}-a-\frac{N}{2}]\) is the zero vector. Hence, we have \(\inf_{x\in[-\frac{a}{2},\frac{a}{2}]} \det G(x)=0\) .

The conditions for \(g\in V_{N,a}\) are technical. We will now give an example, showing that the conclusion about the size of the support of the dual window in Theorem 1.2 may break down if \(g\notin V_{N,a}\) .

$$g(0)=12,\qquad g(-1)=10,\qquad g\biggl(-\frac{4}{3}\biggr)= 5,\qquad g\biggl(- \frac{7}{3}\biggr)=3. $$
$$\begin{aligned} x_{0}-a=-x_{0}-a = -1,\qquad x_{0}- \frac{1}{b}+a=-x_{0}-\frac{1}{b}+a=-\frac {4}{3}, \\ x_{0}-\frac{1}{b}=-x_{0}-\frac{1}{b}=- \frac{7}{3},\qquad x_{0}-\frac{1}{b}-a= -x_{0}- \frac{1}{b}-a =-\frac{10}{3}, \end{aligned}$$
$$\begin{aligned} G(x_{0})= \begin{pmatrix} g(x_{0}+\frac{1}{b}-a) g(x_{0}+\frac{1}{b}) g(x_{0}+\frac{1}{b}+a)\\ g(x_{0}-a) g(x_{0}) g(x_{0}+a)\\ g(x_{0}-\frac{1}{b}-a) g(x_{0}-\frac{1}{b}) g(x_{0}-\frac{1}{b}+a) \end{pmatrix} = \begin{pmatrix} 5 3 0\\ 10 12 10\\ 0 3 5 \end{pmatrix} . \end{aligned}$$
$$G(x)= \begin{pmatrix} 5+ \epsilon_{11}(x) 3+ \epsilon_{12}(x) \epsilon_{13}(x)\\ 10+\epsilon_{21}(x) 12+\epsilon_{22}(x) 10+\epsilon_{23}(x)\\ \epsilon_{31}(x) 3+\epsilon_{32}(x) 5+\epsilon_{33}(x) \end{pmatrix} $$
$$\epsilon_{ij}(x) \rightarrow0 \quad\mbox{as } x\rightarrow x_{0}. $$
$$ \begin{pmatrix} 5+ \epsilon_{11}(x) 3+ \epsilon_{12}(x) \epsilon_{13}(x) \\ 10+\epsilon_{21}(x) 12+\epsilon_{22}(x) 10+\epsilon_{23}(x)\\ \epsilon_{31}(x) 3+\epsilon_{32}(x) 5+\epsilon_{33}(x) \end{pmatrix} \begin{pmatrix} h(x-a) \\ h(x)\\ h(x+a) \end{pmatrix} = \begin{pmatrix} 0 \\ b\\ 0 \end{pmatrix} $$
$$ \begin{pmatrix} 5+ \epsilon_{11}(x) 3+ \epsilon_{12}(x) \epsilon_{13}(x) \\ \eta_{1}(x) \eta_{2}(x) \eta_{3}(x)\\ \epsilon_{31}(x) 3+\epsilon_{32}(x) 5+\epsilon_{33}(x) \end{pmatrix} \begin{pmatrix} h(x-a) \\ h(x)\\ h(x+a) \end{pmatrix} = \begin{pmatrix} 0 \\ b\\ 0 \end{pmatrix} , \quad\mbox{a.e. } x\in\biggl[-\frac{a}{2}, \frac{a}{2} \biggr], $$
$$\begin{aligned} b=\eta_{1}(x)h(x-a)+\eta_{2}(x)h(x) + \eta_{3}(x)h(x+a) \rightarrow0 \end{aligned}$$

On the other hand, the condition \(g\in V_{N,a}\) is not necessary for \(\{ E_{mb}T_{na}g \}_{m,n\in \mathbb {Z}}\) to be a frame in the considered region. For example, let \(N=1\) , \(a=\frac{1}{4}\) , \(b=2\) and take \(g_{1}(x):= (e^{-x^{2}}-e^{-\frac{1}{4}} ) \chi_{[-\frac{1}{2},\frac{1}{2}]}(x)\) . Then elementary calculations show that (A3) does not hold for \(x\in[-\frac{1}{10},-\frac{1}{16}] \subset [-\frac{N}{2},-\frac{N}{4}+\frac{3a}{4}]\) . But since \(\det G(x)>0\) for \(x\in[-\frac{a}{2},\frac{a}{2}]\) , we can prove that \(\{E_{mb}T_{na}g_{1} \}_{m,n\in \mathbb {Z}}\) is a frame by following the steps in the proof of Theorem 1.2 .

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More open than Wineglass Bay, Great Oyster Bay is subject to southerly swells, often providing rougher seas and tall crashing breakers. On reaching the beach, turn right and follow the sand north to the sign posted exit up into the scrub.

For the next 4km the trail hugs the coastline, providing tantalising glimpses of tiny sandy coves and rugged rocks as it slowly makes its way west and then north back towards the car park.

It is no exaggeration to say that this loop would have to be one of the best day walks you will find anywhere in Australia, if not the world!

While the return walk to Wineglass Bay is a favourite of many locals and tourists alike, if you’ve got the time then in our opinion the extension to include Hazards Beach makes for a much more immersive experience, taking you away from the masses and closer to nature.

The initial climb to the saddle is the most strenuous section of the loop, and to be honest it isn’t a slog at all. There are a few small sections north of Hazards Beach that require minimal scrambling, but nothing to be worried about. I’d even go so far as to say that this would be a great introductory ‘long’ walk for children.

All up it took us around 4.5hrs to complete the loop, and that was with a 15 minute break at Wineglass Bay and a 20 minute lunch break at the top end of Hazards Beach.

The weather is generally quite mild on the east coast, but with the warmer conditions it becomes even more important to remember to take plenty of water with you, and to lather up with sunscreen!

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says:

Off to Freycinet for Labour weekend and very much looking forward to doing this walk so thanks for posting – I wasn’t sure if I wanted to do the whole thing but thanks to your detailed info now I know I do. 🙂

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says:

I’m sure you’ll love it Mike!! If you’ve got any questions about the walk or any other aspects of that particular part of Tasmania then just sing out 🙂

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