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We are given an operator $g$ from $\mathcal{l}^2(\mathbb{Z})$ to $\mathcal{l}^2(\mathbb{Z})$, i.e., the space of functions that are square summable over $\mathbb{Z}$ such that $$(g(\phi))(n)=\phi(n+1)+\phi(n-1)-2\phi(n).$$ The question is whether or not $g\geq0$, though the question has multiple parts, some of which may come in useful.
Let $E\in l^2(\mathbb{Z})$ be defined by the equation $E(n)=e^{ink}$. We are asked to find a value $x$ such that $g(E)=xE$. A quick calculation yields $x=(e^{ik}+e^{-ik}-2)$. However, it is obvious that $E$ is not an element of $\mathcal{l}^2(\mathbb{Z})$.

Using the Fourier Transform $\mathcal{F}:L^2([0,2\pi])\rightarrow L^2([0,2\pi])$, where $L^2([0,2\pi])$ is the space of functions that are square integrable over $[0,2\pi]$, we can then put $g$ in momentum representation using $\mathcal{F}^{-1}g\mathcal{F}$, and we get that $$(\mathcal{F}^{-1}g\mathcal{F}\phi)(n)=(e^{ik}+e^{-ik}-2)\sum_{n\in \mathbb{Z}} \langle e^{ink},\phi \rangle e^{ink}.$$

We are then asked to relate $E$ from the first part of the the question to the operator in momentum representation; obviously, the sequence appears in the sum, and if we input an element of the sequence, i.e., $$(\mathcal{F}^{-1}g\mathcal{F}e^{ink})=(e^{ik}+e^{-ik}-2)e^{ink},$$ we get the same value for $x$ in both parts.

We are then asked if $g\geq0$, which is where I'm having trouble. My intuition tells me it's not, as for example, in the momentum representation $\mathcal{F}^{-1}g\mathcal{F}e^{2\pi i}=-4$ and for $e^{ink}$ in both spaces they have the same value for $x$; but I have no idea if this is correct, or if I'm getting the gist of the question. Any help is appreciated.

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  • $\begingroup$ It would be really helpful if you would clearly state the question. First I think the question is "Find $x$ such that $g(e^{ink}) = xe^{ink}$", but then post goes on and I can't tell what you're actually asking. $\endgroup$
    – DanielSank
    Oct 10, 2014 at 4:34
  • $\begingroup$ Sorry, the question had multiple parts. I am fine with that question, as you see above I calculated x. The real problem I am having is determining if g>=0. I put in all the other stuff because it seemed like I had to use the previous results to determine this. $\endgroup$ Oct 10, 2014 at 4:47
  • $\begingroup$ Why did my edit tag on the OP disappear but my edits remain? $\endgroup$
    – DanielSank
    Oct 10, 2014 at 4:49
  • $\begingroup$ Please edit the OP to make it clear what specific question you are asking. $\endgroup$
    – DanielSank
    Oct 10, 2014 at 4:54

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You found $$x = e^{ik} + e^{-ik} - 2 = 2(\cos(k) - 1) \leq 0 . $$ If the question is whether $g\ge 0$ then the answer is "no" in the sense that it's eigenvalues are non-positive as we have just showed.

A bit of intuition

We recognize $$\phi(n+1) + \phi(n-1) -2\phi(n)$$ as some kind of second derivative. Consider the first derivative of a function $f$ (taking limits from the right) $$\left. \frac{df}{dx}\right|_{x_0} = \lim_{h\rightarrow 0} \frac{f(x_0+h) - f(x_0)}{h}.$$ Now take the derivative of that (using limits from the left) $$\frac{\frac{f(x_0+h) - f(x_0)}{h} - \frac{f(x_0)-f(x_0-h)}{h}}{h} = \frac{f(x+h)+f(x_0-h) - 2f(x_0)}{h^2}$$ Of course, in the discrete case the $h$'s go away, but you can see the structure of this second derivative is exactly the same as the $g$ operator. So, we can immediately guess that the answer is $x=-k^2$, which is non-positive.

What's the relation between the discrete and continuous cases? Well, in the discrete case everything is periodic and you get the $\cos$ function. The continuous case is like the discrete case where the points are so close together than no matter how far you go you're always still close to 0. Therefore, you only ever feel the first non-constant term in the Taylor series of the $\cos$, namely the parabolic function. This is just a heuristic, but it's a good heuristic.

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  • $\begingroup$ Awesome, thank you very much! The operator was actually called the "discrete Laplacian" in the textbook, and this makes a lot of sense from your explanation. Funny that it didn't mention any of this. Thanks again. $\endgroup$ Oct 10, 2014 at 5:22

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