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Information needed

$$I = (a,b)$$

$$ H_0^1 (I) = \{ v \in H^1(I) : v(a)= v(b) = 0 \} $$

$$ H^1 (I) = \{ v: v , v' \in \mathbb{L}_2(I) \} $$

$$ \mathbb{L}_2(I) = \{ v:v \text{ it's defined at } I \text{ and } \int v^2\,dx < \infty \} $$

Prove that this is a scalar product in $H^1_0 $

Using the notation $$ \langle u,v\rangle = \int_\Omega [ \bigtriangledown u \cdot\bigtriangledown v +uv ] $$

where $\Omega$ is a bounded domain.

In the book numerical solutions of PDE by the finite element method by Claes there is this passage and I could not see clearly this statement.

Thanks to any help !

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    $\begingroup$ Have you tried proving it yourself? An inner product has three defining properties, each of which should not be too hard to verify. $\endgroup$ – MaoWao Jan 30 at 12:15
  • $\begingroup$ you mean, symmetry; linearity and podsitive definiteness ? $\endgroup$ – Saiten Jan 30 at 12:19
  • $\begingroup$ It's bilinearity (linearity in each argument separately), not linearity. Otherwise yes. $\endgroup$ – MaoWao Jan 30 at 12:23
  • $\begingroup$ Use $\langle X\rangle$ for $\langle X\rangle$. $\endgroup$ – Shaun Jan 30 at 12:39
  • $\begingroup$ Thankss for the help! $\endgroup$ – Saiten Jan 30 at 13:24
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Once you establish that the expression is well defined, you just need to verify the conditions for an inner product.

  1. $\langle u,v \rangle = \langle v,u \rangle$.
  2. $\langle a u, v\rangle = a \langle u,b\rangle$ and $\langle u+v,w\rangle = \langle u,w\rangle+\langle v,w\rangle$.
  3. $\langle u,u\rangle\ge 0$ and $\langle u,u\rangle=0 \Leftrightarrow u = 0$

The third property is the trickiest. You need to show that if $u \in H^1_0$ and $$ \int_I (|\nabla u|^2 + u^2) = 0 $$ then $u=0$ almost everywhere in $I$.

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  • $\begingroup$ Use $\langle X\rangle$ for $\langle X\rangle$. $\endgroup$ – Shaun Jan 30 at 13:17

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