For Complete Ritz Sequence Continuous Functional approaches its Minimal Value
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Theorem
Let $J$ be a continuous functional.
Let $\sequence {\phi_n}$ be a complete Ritz sequence.
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Then:
- $\ds \lim_{n \mathop \to \infty} \mu_n = \mu$
where $\ds \mu = \inf_y J \sqbrk y$.
Proof
Let $y^*: \R \to \R$ be such that:
- $\forall \epsilon > 0: J \sqbrk {y^*} < \mu + \epsilon$
By assumption of continuity of $J$:
- $\forall \epsilon > 0: \exists \map \delta \epsilon > 0: \paren {\size {y - y^*} < \delta} \implies \paren {\size {J \sqbrk y - J \sqbrk {y^*} } < \epsilon}$
Let $\eta_n = \bsalpha \bsphi$, such that:
- $\exists n \in\ N: \exists N \in \N: \paren {n > N} \implies \paren {\size {\eta_n - y^*} < \epsilon}$
where $\bsalpha$ is an $n$-dimensional real vector.
Let $y_n = \bsalpha \bsphi$, where $\bsalpha$ is such that it minimises $J \sqbrk {y_n}$.
Hence:
- $J \sqbrk {y_n} < J \sqbrk {\eta_n} + \epsilon < \mu + 2 \epsilon$
On the other hand, by the definition of $y_n$:
- $\mu \le J \sqbrk {y_n} \le J \sqbrk {\eta_n}$
Thus:
- $\mu \le J \sqbrk {y_n} < \mu + 2 \epsilon$
If $J \sqbrk {y_n} = \mu$, then the inequality is satisfied, and $n$ is finite.
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If $ \mu < J \sqbrk {y_n}$, then subtract $\mu$ from all the terms:
- $0 < J \sqbrk {y_n} - \mu < 2 \epsilon$
We have that:
- $\paren {\epsilon > 0} \implies \paren {-2 \epsilon < 0}$
Hence:
- $-2 \epsilon < J \sqbrk {y_n} - \mu < 2 \epsilon$
or:
- $\size {J \sqbrk {y_n} - \mu} < 2 \epsilon$
Since $\epsilon$ and $n$ inherits their constraints from the definitions of $\eta_n$ and the continuity of $J$, these together with the last inequality imply:
- $\ds \lim_{n \mathop \to \infty} J \sqbrk {y_n} = \mu$
where $J \sqbrk {y_n} = \mu_n$.
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 8.40 $: The Ritz Method and the Method of Finite Differences