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 $H_0^1(U)$?

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1 Answer 1


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 $(-1,1)\times(0,1)$.

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


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

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!

  • $\begingroup$ 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 $H_0^1$ $\endgroup$
    – Simmetrico
    Dec 14, 2022 at 10:02

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