Show two functions are uniformly comparable on the real line I am trying to find inequalities between this two functions $ f(k)=| (\lambda^2+k^2)^{\alpha/2}\cos ( \alpha \arctan \frac{|k|}{\lambda} )-\lambda^{\alpha} |$ and $ g(k)=|k|^{\alpha} $  such as, $$ c_1 g(k)\le f(k)\le c_2 g(k),$$ where $ 0<c_1<c_2 $, $ k \in \mathbb{R} $, and the parameters $\alpha$ and $\lambda $ fixed and $ 1<\alpha<2 $, $ \lambda>0 $.  
Of course, the constants $ c_1,c_2 $ can be dependent to the parameters $ \alpha $ and $ \lambda $, but cannot be dependent to the variable $k$.
Is there any chance to find an inequality something like that.
 A: The ratio of $f(k)$ to $g(k)$ can become arbitrarily small as $k\to 0$.  Here's one way of showing that.
First let us make the substitution $z = \frac{\lambda}{|k|}$, so where $k \neq 0$, we have $z \in (0,+\infty)$.  Note that as $|k| \to 0$, $z \to +\infty$, and conversely.
Then:
$$ \frac{f(k)}{g(k)} = \left| \left(\frac{\lambda^2}{|k|^2} + 1 \right)^{\alpha/2}
 \cos\left(\alpha \cot^{-1} \frac{\lambda}{|k|}\right) - \left(\frac{\lambda}{|k|}\right)^\alpha \right| $$
becomes:
$$ r(z) \equiv |(z^2+1)^{\alpha/2} \cos(\alpha \cot^{-1} z) - z^\alpha | $$
Let's make the further substitution $x = \cot^{-1} z$, so where $z \in (0,+\infty)$, we get $x \in (0,\pi/2)$.  Note that as $z \to +\infty$, $x \to 0^+$, and conversely.  So in terms of the original variable $k$, $|k|$ approaches zero if and only if $x$ approaches zero from above.  The substitution yields $r(z) = |R(x)|$ where by trigonometric identities:
$$ R(x) = \cot^\alpha x - \csc^\alpha x \cos(\alpha x) \\
  = \frac{\cos^\alpha x - \cos(\alpha x)}{\sin^\alpha x } $$
Properties of R(x)
The function $R(x)$ on $(0,\pi/2)$ has the following properties, assuming as in the Question that $1 \lt \alpha \lt 2$:


*

*$R(x)$ tends to zero as $x \to 0^+$.

*$R(x)$ is strictly increasing on $(0,\pi/2)$.

*$R(x)$ tends to $R(\pi/2)$ as $x$ tends to $\pi/2$.


The last of these facts follows easily from the continuity of $R(x)$ from the left at $\pi/2$, where $\sin(x) \neq 0$. Since $\cos(\pi/2) = 0$, $\sin(\pi/2) = 1$, note $R(\pi/2) = -\cos(\alpha \pi/2) \gt 0$ as the angle $(\alpha \pi/2)$ is in the second quadrant (because $1\lt \alpha \lt 2$).
The proof of 1. is also dependent on $1\lt \alpha \lt 2$, as it amounts to showing the numerator of $R(x)$ tends to zero like $O(x^2)$, while the denominator tends to zero like $x^\alpha$.  Thus $\alpha \lt 2$ compels $R(x)$ to approach zero.
The numerator $\cos^\alpha x - \cos(\alpha x)$ is analytic at $x=0$, i.e. it has a power series expansion about $x=0$ valid up to the nearest singularity (e.g. at $x=\pi/2$ where cosine vanishes).  Thus it suffices to show the constant term and first derivative of the numerator vanish, so that the power series has leading term $O(x^2)$ as claimed.
Clearly the constant term is zero since $\cos 0 = 1$.  Differentiating the numerator gives:
$$ \alpha (-\cos^{\alpha-1} x \sin x + \sin(\alpha x) ) $$
Since $\sin 0 = 0$, this first derivative vanishes as well at $x=0$.  This establishes that no positive constant $c_1$ exists s.t. $c_1 g(k) \le f(k)$.
Monotonicity
To show that $R(x)$ is strictly increasing on $[0,\pi/2]$, it suffices to show that its derivative is positive on $(0,\pi/2)$.  Differentiating its first expression:
$$ R'(x) = \alpha (-\csc^2 x \cot^{\alpha - 1} x + \csc^\alpha x \cot x \cos \alpha x
    + \csc^\alpha x \sin \alpha x ) \\
    = \alpha \csc^{\alpha + 1} x (-\cos^{\alpha - 1} x + \cos x \cos \alpha x
    + \sin x \sin \alpha x ) \\
    = -\alpha \; \frac{\cos^{\alpha - 1} x - \cos (\alpha - 1) x}{\sin^{\alpha + 1} x}
$$
Note the similarity of $R'(x)$ to our second expression for $R(x)$, particularly their numerators.  This similarity motivated a separate Question about the monotonicity of that family of functions, where the order of subtraction is reversed.
In any case it suffices to show numerator $N(x) = \cos^{\alpha - 1} x - \cos (\alpha - 1) x$ is negative on $(0,\pi/2)$ in order to prove $R'(x)$ is positive there, since $\sin x \gt 0$ there.  Since $N(0)= 0$ with right continuity, it is enough to show $N(x)$ is strictly decreasing on $(0,\pi/2)$, just as Julián Aguirre shows in the first case of his Answer on that problem.  Replacing $\alpha$ by $\alpha-1$ in our earlier differentiation of the numerator:
$$ N'(x) = (\alpha - 1) (-\cos^{\alpha-2} x \sin x + \sin (\alpha-1) x) ) $$
Since $0 \lt \alpha - 1 \lt 1$, we have $0 \lt \sin (\alpha - 1) x \lt \sin x$ on $(0,\pi/2)$:
$$ N'(x) \lt (\alpha - 1) (\sin x) (-\cos^{\alpha-2} x + 1 ) $$
But $\alpha - 2 \lt 0$, so $\cos^{\alpha-2} x \gt 1$ and $N'(x)$ is negative on $(0,\pi/2)$.
Summary
If $|k|$ approaches infinity, the ratio $f(k)/g(k)$ increases, approaching $R(\pi/2)$ from below.  Therefore the best constant $c_2$ is $R(\pi/2)$.
For $k$ over the entire real line there is no positive lower bound on the ratio.  However if $k$ were bounded away from zero, say restricting $|k| \ge k_0 \gt 0$, then a suitable constant $c_1$ could be devised from the above analysis.
