I want to prove a well-known fact in $L^p(R^n)$ namely that, the convolution of an element in $L^p$ with an element of $L^1$ is in $L^p$

1. $u∈L^p(R;X),f∈L^1(R)$ So then both $u$ and $f$ are strongly measurable $⇒u(τ)f(t-τ)$ is strongly measurable. X Separable and $u(τ)f(t-τ)$ strongly measurable Implies that $u(τ)f(t-τ)$ is strongly integral. Notice that the integral need not be finite, in what follows I seek to prove that it's "p" power has indeed finite integral

2. Using the fact that linear forms commute with the integral I get that: $∀ h∈L^{p^*}(R;X^*): ⟨h(t),∫_R u(t-τ)f(τ)dτ⟩=∫_R⟨h(t),u(t-τ)f(τ)⟩dτ$. Since $f∈L^1 (R), f$ will behave like a scalar and :$∫_R⟨h(t),u(t-τ)f(τ)⟩dτ=∫_Rf(τ)⟨h(t),u(t-τ)⟩dτ≤∫_R|f(τ)|‖h(t)‖_{X^*} ‖u(t-τ)‖_X dτ$

3. We finally prove that $(u*f)(t)∈L^p (R;X)$. Let $h∈L^{p^*}(R;X^*)$. The duality relationship between $h$ and $(u*f)$ is $∫_R⟨h(t),(u*f)(t)⟩dt$. using (2) I get $∫_R⟨h(t),(u*f)(t)⟩dt≤∫_R∫_R|f(τ)| ‖h(t)‖_{X^*} ‖u(t-τ)‖_X dτdt$. All the functions are real and measurable, so we can change the order of integration and with Holder : $∫_R⟨h(t),(u*f)(t)⟩≤∫_R|f(τ)| ‖h‖_{p^*,X^*} ‖u‖_{p,X} dτ≤‖f‖_{1}‖h‖_{p^*,X^*} ‖u‖_{p,X}≤c(f,u)‖h‖_{p^*,X^*} $

4. So $(u*f)∈(L^{p^*}(R;X^*))^*=L^{p}(R;X)$ since X reflexive.

Is this proof correct ? If not can you please tell me what I should change. In addition is my result accurate ? Meaning that my assumptions on X are absolutely necessary ?