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Question about Evans states, chapter 8.4.2!

We have $I[w] := \int_U \frac{1}{2}|Dw|^2 - fw\, dx$, among all functions $w$ belonging to the set

$$\mathcal{A} : = \{w \in H_0^1(U) : w \geq h \, \mbox{ a.e. in } U\}$$

with smooth $h$ and $f$.

And we have:

Theorem: Assume the admissible set $\mathcal{A}$ is nonempty. Then there exists a unique function $u \in \mathcal{A}$ satisfying

$$I[u] = min_{w \in \mathcal{A}} I[w]$$

I don't understand how we can say that $L$ is weakly lower semicontinuous. And I think we need to have this for the proof of the existence.

Daniela
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  • Aside the constraint condition, there is this result that says that if your integrand is convex with respect to the gradient argument and your integrand has a lower bound, then the integral functional is weakly lower semi-continuous. – Chee Han Jun 07 '15 at 00:39
  • I know this result. But in this case, these conditions are checked? – Daniela Jun 07 '15 at 09:23
  • Yea, $x\mapsto x^2$ is convex, and the coercivity estimate isn't too difficult to establish I think, using Poincare's inequality for the first one and Cauchy-Schwarz plus Young's inequality for the second one probably? – Chee Han Jun 07 '15 at 09:37
  • we need to see that $\frac{1}{2} |Dw|^2 - fw$ is lower bounded, right? I don't know how I can use Poincare's inequality, because I don't have any norm :/ – Daniela Jun 07 '15 at 09:46
  • Yes, Poincare's inequality tells you that the $L^2$ norm of the gradient is equivalent to the full $H^1$-norm, so you can bound the first term below by the $H^1$ norm. For the second term, apply Cauchy-Schwarz, then use Young's inequality (some called this inequality by other name but my lecturer always calls it this way) $2ab\le a^2/2\varepsilon + \varepsilon b^2/2$. Then choose $\varepsilon$ so that the constant associated to the $H^1$ norm of w is positive. – Chee Han Jun 07 '15 at 09:59
  • Ahh thanks :) My problem was understand the concept of bounded here – Daniela Jun 07 '15 at 10:06
  • @CheeHan Be careful: the integrand does not have a lower bound, only the integral does. –  Jun 08 '15 at 02:56
  • Absolutely, but all it matters is the coercivity estimate :p – Chee Han Jun 08 '15 at 08:58

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