If $u$ is in a Sobolev space on $U$, then $u$ is in Sobolev space on $V \subset U$

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Let $u \in W^{k,p}(U)$ where $U = (a,b)$ is an open interval in $\mathbb{R}$. Let $V \subset U$ be open, then how to prove that $u \in W^{k,p}(V)$?


Attempt:

Firstly, if $u$ is locally summable in $U$, then of course $u$ is also locally summable in $V$.

And we already know that there exist a function $v$ that satisfies

$$ \int_{U} u D^{\alpha} \phi \:dx = \int_{U} v \phi \:dx $$

for all $\phi \in C_{c}^{\infty}(U)$.

How to prove there exist a function $w$ that satisfies

$$ \int_{V} u D^{\alpha} \phi \:dx = \int_{V} w \phi \:dx $$

for all $\phi \in C_{c}^{\infty}(V)$

Intuitively, the functions $w$ is apparently the function $v$ evaluated only along $V \subset U$. But how to show this?