Let $f \in L^p(\mathbb{R}), p\ge1.$ $\omega_k(f;t)_p,$ the $k^{th}$ order modulus of smoothness of $f,$ is defined by $$\omega_k(f;t)_p = \sup_{0<|h| \le t} \|\Delta_h^kf\|_p,$$ where $$\Delta_h^kf(x) = \sum_{r=0}^k (-1)^{k-r} \binom{k}{r}f(x+rh).$$
For $\alpha>0, 1\le p, q <\infty,$ we define the Besov space $B_q^\alpha(L^p)$ as following. For $f\in L^p$ if the quantity $$|f|_{B_q^\alpha(L^p)} = \left(\int_0^\infty \left(\dfrac{\omega_k(f;t)_p}{t^\alpha}\right)^q\dfrac{dt}{t}\right)^{1/q}$$
is finite, where $k=\lfloor\alpha\rfloor + 1,$ then $f \in B_q^\alpha(L^p).$ Please note that $|f|_{B_q^\alpha(L^p)}$ is a seminorm. This book and this article claim that the integral $\int_0^\infty$ in the seminorm $|f|_{B_q^\alpha(L^p)}$ can equivalently be replaced by $\int_0^1.$ One way inequality is very simple but I find it very difficult to prove other way inequality i.e.
$$\left(\int_0^\infty \left(\dfrac{\omega_k(f;t)_p}{t^\alpha}\right)^q\dfrac{dt}{t}\right)^{1/q} \le C\left(\int_0^1 \left(\dfrac{\omega_k(f;t)_p}{t^\alpha}\right)^q\dfrac{dt}{t}\right)^{1/q}, C>0.$$
I tried using the relation $\omega_k(f;t)_p \le 2^k\|f\|_p$ but was unable to prove. I suspect that there is some fundamental fact which I am missing. May be the fact $\|f\|_p<\infty$ tells something that I am unable to see. Can someone please help me.
PS: I have seen this question but it does not provide an answer for my question.