For Complete Ritz Sequence Continuous Functional approaches its Minimal Value

Let $J$ be a continuous functional.

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

Then:


 * $\displaystyle\lim_{n\mathop\to\infty}\mu_n=\mu$

where $\displaystyle\mu=\inf_y J\sqbrk y$

Proof
Let $y^*:\R\to\R$ be such that:


 * $\displaystyle\forall\epsilon>0:J\sqbrk{y^*}<\mu+\epsilon$

By assumption of continuity of $J$:


 * $\displaystyle\forall\epsilon>0:\exists\map {\delta} {\epsilon}>0:\paren{\size {y - y^*}<\delta}\implies\paren{\size {J\sqbrk y-J\sqbrk{y^*} }<\epsilon}$

Let $\eta_n=\boldsymbol\alpha\boldsymbol\phi$, such that:


 * $\displaystyle\exists n\in\N:\exists N\in\N:\paren{n>N}\implies\paren{\size {\eta_n-y^*}<\epsilon}$

where $\boldsymbol\alpha$ is an $n$-dimensional real vector.

Let $y_n=\boldsymbol\alpha\boldsymbol\phi$, where $\boldsymbol\alpha$ is such that it minimises $J\sqbrk{y_n}$

Hence:


 * $\displaystyle J\sqbrk{y_n}0}\implies\paren{-2\epsilon<0}$

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:


 * $\displaystyle\lim_{n\mathop\to\infty} J\sqbrk{y_n}=\mu$

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