# Mollifiers and Banach space valued functions

In Section 5.9.2 of Evans PDE, it gives some properties of Sobolev space involving time.

Suppose $$u\in L^2(0,T;H_0^1(U))$$. Extend $$u$$ to be $$0$$ on $$(-\infty,0)$$ and $$(T,\infty)$$ and then set $$u^\varepsilon=\eta_\varepsilon\ast u$$, $$\eta_\varepsilon$$ denoting the usual mollifier on $$\mathbb{R}^1$$. In the proof of Theorem 3, it says that "Fix any point $$s\in (0,T)$$ for which $$u^\varepsilon(s)\rightarrow u(s) \text{ in }L^2(U)."$$ My questions are:

1. why can we find such $$s$$? Since $$u^\varepsilon\rightarrow u$$ in $$L^2(0,T;H_0^1(U))$$, it seems that we can only get a subsequence of $$u^\varepsilon$$ converging to $$u$$ almost everywhere?

2. Why consider convergence in $$L^2(U)$$ not in $$H_0^1(U)$$? Maybe it is too later, but I meet an analogous problem with the same proof.

The point 1. is correct as you say: we can find a subsequence such that converges almost everywhere.

For point 2. I think that we consider the Sobolev space $$H^{1}_0(U)$$ instead of $$L^2(U)$$ is also correct.

I add a question, because I don't understand the passage from the scalar product in $$L^2(U)$$ and the duality product under the integral. I think that in the hypothesis (ii) one must replace replace $$L^2(U)$$ with $$H^{1}_0(U)$$, otherwise the thesis is false. As counterexample one can consider the function $$u(x,t)=t(1-x^2)$$ defined in $$(-1,1)\times(0,1)$$.

$$\frac{d}{dt}\|u(t)\|_{L^2(-1,1)}^2=\frac{d}{dt}\int_{-1}^1 t^2(1-x^2)^2dx=2t\int_{-1}^1 (1-x^2)^2dx,$$

while

$$2\langle u'(t),u(t)\rangle=2\int_{-1}^1(1-x^2)t(1-x^2)dx+2\int_{-1}^1 4x^2t.$$

Note that I suppose that $$\langle\cdot,\cdot\rangle$$ is the duality product between $$H^{-1}$$ and $$H^1_0$$.

Moreover, in the last part of the proof that you posted there is an estimate that follows immediately if one considers the setting I proposed. And in this case the thesis obtained is more stronger.

I accept any hint or correction!

• I think that the reasoning below isn't correct because one can identify $\langle\cdot,\cdot\rangle$ whit $(\cdot,\cdot)$ (inner product in $L^2$) if $u$ and $u'$ are enough regular, in the sense that $\langle\cdot,\cdot\rangle$ does not coincide whit the inner product in $H_0^1$ Dec 14, 2022 at 10:02