# Equivalent Besov seminorm - change of integral limits

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.