# For Complete Ritz Sequence Continuous Functional approaches its Minimal Value

## Theorem

Let $J$ be a continuous functional.

Let $\sequence {\phi_n}$ be a complete Ritz sequence.

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.

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 $J \sqbrk {y_n} = \mu_n$.

$\blacksquare$