I will solve the problem below by $1)$ working in reciprocal (wavenumber space) and in $2)$ energy space ($\epsilon$). But first some contextual background:
Graphene is a single sheet of carbon atoms which is a two-dimensional metal. Similarly, there are metallic surface states in $\mathrm{Bi_2Se_3}$ which are insulating in the bulk. These mobile two-dimensional electron systems have an unconventional energy dispersion near an energy $\epsilon_0$. There are two energy bands of electron states which touch at $\epsilon_0$. The lowest state of the upper band is at $\epsilon_0$, and so is the highest state of the lower band. Near $\epsilon_0$, the states in the upper band have the dispersion relation: $\epsilon_{\bf{k}}^{+}=\epsilon_0+\hbar\,v_F\lvert\bf{k}\rvert$ for $\epsilon_{\bf{k}}\gt \epsilon_0$. The electrons in the lower band have the dispersion relation $\epsilon_{\bf{k}}^{-}=\epsilon_0 - \hbar\,v_F\lvert \bf{k}\rvert$ for $\epsilon_{\bf{k}}\lt \epsilon_0$.
Show that the density of states (per spin state) of the upper band near the Fermi level is given by:$$g^{+}(\epsilon)=\frac{A}{2\pi\hbar^2v_F^2}\left(\epsilon-\epsilon_0\right)\tag{1}$$ for $\epsilon\gt \epsilon_0$ and zero otherwise. ($A$ is the area of the sample.)
To solve this I start from the definition of number of states: $$\delta N_{\mathrm{States}}=g(\epsilon)\delta\epsilon$$
$$g(\epsilon)\delta\epsilon=g(k)\delta k=\frac{A}{\left(2\pi\right)^2}\times 2\pi k\delta k=\frac{A}{2\pi}\times k\delta k\tag{2}$$
where I have considered an annulus of inner radius $k$ and outer radius $k+\delta k$; which means that the thickness of the 'ring' (blue in diagram) is $\delta k$
From the dispersion relation for $\epsilon_{\bf{k}}\gt \epsilon_0$, $$k=\frac{\left(\epsilon-\epsilon_0\right)}{\hbar v_F}$$ Substituting into $(2)$, $$g(\epsilon)\delta\epsilon=\frac{A}{2\pi}\frac{\left(\epsilon-\epsilon_0\right)}{\hbar v_F}\delta k\tag{3}$$ From derivative definition, $$\delta k=\frac{d k}{d \epsilon}\delta \epsilon=\frac{1}{\hbar v_F}\delta\epsilon$$ Substituting into $(3)$ gives, $$g(\epsilon)\delta\epsilon=\frac{A}{2\pi}\frac{\left(\epsilon-\epsilon_0\right)}{\left(\hbar v_F\right)^2}\delta\epsilon$$ gives the required result, $$g^{+}(\epsilon)=\frac{A}{2\pi\hbar^2v_F^2}\left(\epsilon-\epsilon_0\right)$$ for $\epsilon\gt \epsilon_0$ and zero otherwise.
Okay, fine, so that was in $k$-space, now I should be able to get the same exact same result by working in $\epsilon$-space (energy space):
So as before, $$\delta N_{\mathrm{States}}=g(\epsilon)\delta\epsilon=\frac{A}{\left(2\pi\right)^2}\times 2\pi \epsilon\delta\epsilon\tag{4}$$ where $2\pi \epsilon\delta\epsilon$ is the area of the annulus in energy space with inner radius $\epsilon$ and outer radius $\epsilon+\delta\epsilon$, the thickness of the annulus is $\delta\epsilon$. From $(4)$, I immediately see that the density of states is, $$g^{+}(\epsilon)=\frac{A}{2\pi}\epsilon$$ which is not the same as $(1)$. Now, it was my understanding that when finding the density of states I have a choice of which space to work in (energy space or reciprocal space); The result should be independent of the representation I choose to calculate it in.
These 2 results must be the same, so how can the results be different? Or, put in another way, how can $$\frac{A}{2\pi}\epsilon=\frac{A}{2\pi\hbar^2v_F^2}\left(\epsilon-\epsilon_0\right)?$$
I voted to migrate this question to physics.SE, since in my view it’s more about physics than about math, but since it’s here now and it doesn’t look like it’s going anywhere, I might as well answer it.
Despite your insistence using various forms of emphasizing font modifications that the results should be the same, it would actually be surprising if they were.
$g(k)\delta k=\frac A{\left(2\pi\right)^2}2\pi k\delta k$ expresses the assumption that the density of states is constant in $2$-dimensional $k$-space, since $2\pi k\delta k$ is just the area of the annulus in $k$-space. This assumption can be physically derived by considering the system in a box and letting the dimensions of the box go to infinity.
By contrast, $g(\epsilon)\delta\epsilon=\frac A{\left(2\pi\right)^2}2\pi\epsilon\delta\epsilon$ expresses no physical assumption. It would correspond to the assumption that the density of states is constant in some $2$-dimensional energy space. There’s no such space (energy is a scalar quantity, not a vector), and no reason why the density of states should be constant with respect to the energy. In fact, the entire question would be trivial if this relation were correct, as there’s no need to calculate the density of states per energy from the dispersion relation if you’re already assuming that it’s fixed.