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I am referring to the book Introduction to Functional Analysis to Boundary Value Problems and Finite Element by Daya Reddy (page 233-234) for this question.

I am just stuck between parts of Poincare's Inequality proof as explained in the book. The statement is:

For any $u$ $\in$ $H^{k}(\Omega)$ where, $\Omega$ is $\mathbb{R} ^n$ with a smooth boundary (Lipschitz Continous) there exists constants $c_1$ & $c_2$ s.t. $$\|u\|_{H^k}^2 \le c_1 \int_\Omega \sum_{|\alpha| = k} \left(\mathscr{D}^\alpha u\right)^2 + c_2 \sum_{|\alpha| < k} \left( \int_\Omega \mathscr{D}^\alpha u \text{ } dx \right)^2 $$

where $\mathscr{D}$ is the differential operator, $H^k(\Omega) := \{f: \text{ } \mathscr{D}^\alpha f \text{ } \in \text{ } \mathscr{L}^2 (\Omega) \quad \forall \text{ } \alpha \textit{ such that } | \alpha | \le k \}$

For $ \quad n = 1 \Rightarrow \Omega = (a, b)$, we can write: $$\|u\|_{H^k}^2 \le c_1 \int_a^b \left(\frac{d^ku}{dx^k}\right)^2dx \quad + \quad c_2 \sum_{j < k} \left(\int_a^b\frac{d^ju}{dx^j}\right)^2dx $$

For $\quad (\eta , \xi) \quad in \quad (a, b) \quad | \quad \eta < \xi$ $$ [u(\xi) - u(\eta)] = \int_\eta^\xi u' dx \quad \Rightarrow \quad [u(\xi) - u(\eta)]^2 = \left(\int_\eta^\xi u' dx \right)^2 $$ By Cauchy-Schwarz inequality, $$[u(\xi) - u(\eta)]^2 = \left(\int_\eta^\xi (1)(u') dx \right)^2 \le \left(\int_\eta^\xi 1^2 dx \right)^2 \left(\int_\eta^\xi (u')^2 dx \right) \\ \le (b - a)\int_a^b(u')^2 dx $$

Now, we integrate with respect to $\xi $, keeping $\eta$ constant, and then with respect to $\eta$ to get

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$$ (b-a)\left[\int_a^b u^2 (\xi) d\xi + \int_a^b u^2 (\eta) d\eta \right] - 2 \int_a^b u(\xi) d\xi \int_a^b u(\eta) d\eta \\ \le (b-a)^3 \int_a^b (u')^2 dx$$

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I have been working on how we get to the expression (above). Please provide any suggestion you can.

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I think you probably mean $\xi > \eta$?

The overall inequality is $$ |u(\xi) - u(\eta)|^2 \leq (b-a)\int_a^b (u')^2~dx. $$ Write this as $$ u(\xi)^2 - 2 u(\xi)u(\eta) + u(\eta)^2 \leq (b-a)\int_a^b (u')^2~dx. $$ Integrate in $\xi$ from $a$ to $b$. $u(\eta)^2$ does not depend on $\xi$, so we obtain a factor of $(b-a)$. Similarly with the integral on the right. Everything is nonnegative so we preserve the direction of the inequality. So we obtain $$ \int_a^b u(\xi)^2~d\xi - 2 u(\eta)\int_a^b u(\xi)~d\xi + (b-a)u(\eta)^2 \leq (b-a)^2\int_a^b (u')^2~dx. $$ Now do the same, but with respect to $\eta$: $$ (b-a)\int_a^b u(\xi)^2~d\xi - 2 \int_a^b u(\eta)~d\eta\int_a^b u(\xi)~d\xi + (b-a)\int_a^b u(\eta)^2~d\eta \leq (b-a)^3\int_a^b (u')^2~dx. $$ And we're done!

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  • $\begingroup$ I can't believe the ans was so simple. Thanks !!! $\endgroup$ – DOOM Nov 23 '14 at 5:05

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