2 days ago, by one potato two potato
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$\newcommand{\Q}{\mathbb{Q}}$PoincarΓ© duality tells you that there are non-degenerate pairings $H^i(M) \otimes H^{n-i}(M, \partial M) \to \Q$ for all $0 \le i \le n$.
Using the long exact sequence of the pair $(M, \partial M)$, the known facts that $H^n(M) = 0$ and $H^n(M,\partial M) = \Q$ (Poin...
0
A necessary condition for the differentiable functional $J[y]$ to have an extremum for $y=\hat{y}$ is that variation vanish for $y=\hat y$, i,e., that
$$\delta J[h]=0$$ for $y=\hat y$ and for all admissible $h$.
I was referring the following proof given in The Textbook Calculus of Variation, Gel...
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Some copious hints:
$1).$ It is enough to show that every open covering of $\mathbb R_{\ell}$ by basis elements has a countable subcover.
$2).$ Let $\mathscr A = \{[a_i, b_i) | i β J\}$ be such a cover and consider the union of intervals $A' = \bigcup_{i\in J} (a_i, b_i).$
$3).$ If $x\in \math...
My Doubt
Here $||f||_{1}=\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|$ where
as $||f||_0=\sup_{x\in[0,1]}|f(x)|$. We can easily prove from
definition that
$$||f||_0=\sup_{x\in[0,1]}|f(x)|\leq
\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|=||f||_1$$
Using the above inequality, ...
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