Let $\Omega$ be an open set from $\mathbb{R}^N$ and $X=L^2(\Omega)$. How can we prove that if $u,v\in H^1(0,T,X)$ (in Bochner sense) then $(u\cdot v)'\in L^2(0,T,X)$ with $(uv)'=u'v+v'u$ and the integration by parts formula holds:

$$\int_0^T u'v\ dt=u(T)v(T)-u(0)v(0)-\int_{0}^T uv'\ dt $$

?

I see this formula applied in this context, but I found no proof for it.

Let $\Omega$ be an open set from $\mathbb{R}^N$. How can we prove that if $u,v\in H^1(0,T,L^2(\Omega))$ (in Bochner sense) then $(u\cdot v)'\in L^2(0,T,L^1(\Omega))$ with $(uv)'=u'v+v'u$ and the integration by parts formula holds:

$$\int_0^T u'v\ dt=u(T)v(T)-u(0)v(0)-\int_{0}^T uv'\ dt $$

?

I see this formula applied in this context, but I found no proof for it.

Let $\Omega$ be an open set from $\mathbb{R}^N$ and $X=L^2(\Omega)$. How can we prove that if $u,v\in H^1(0,T,X)$ (in Bochner sense) then $u\cdot v\in H^1(0,T,X)$ with $(uv)'=u'v+v'u$ and the integration by parts formula holds:

$$\int_0^T u'v\ dt=u(T)v(T)-u(0)v(0)-\int_{0}^T uv'\ dt $$

?

I see this formula applied in this context, but I found no proof for it.

Let $\Omega$ be an open set from $\mathbb{R}^N$ and $X=L^2(\Omega)$. How can we prove that if $u,v\in H^1(0,T,X)$ (in Bochner sense) then $(u\cdot v)'\in L^2(0,T,X)$ with $(uv)'=u'v+v'u$ and the integration by parts formula holds:

$$\int_0^T u'v\ dt=u(T)v(T)-u(0)v(0)-\int_{0}^T uv'\ dt $$

?

I see this formula applied in this context, but I found no proof for it.

Let $\Omega$ be an open set from $\mathbb{R}^N$ and $X=L^2(\Omega)$. How can we prove that if $u,v\in H^1(0,T,X)$ (in Bochner sense) then $u\cdot v\in H^1(0,T,X)$ with $(uv)'=u'v+v'u$ and the integration by parts formula:

$$\int_0^T u'v\ dt=u(T)v(T)-u(0)v(0)-\int_{0}^T uv'\ dt $$

?

I see this formula applied in this context, but I found no proof for it.

Let $\Omega$ be an open set from $\mathbb{R}^N$ and $X=L^2(\Omega)$. How can we prove that if $u,v\in H^1(0,T,X)$ (in Bochner sense) then $u\cdot v\in H^1(0,T,X)$ with $(uv)'=u'v+v'u$ and the integration by parts formula holds:

$$\int_0^T u'v\ dt=u(T)v(T)-u(0)v(0)-\int_{0}^T uv'\ dt $$

?

I see this formula applied in this context, but I found no proof for it.