For Complete Ritz Sequence Continuous Functional approaches its Minimal Value

Let $ J $ be a continuous functional.

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

Then:


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

where $ \displaystyle \mu = \inf_y J \left [ { y } \right ] $

Proof
Let $ y^* $ be such that:


 * $ \displaystyle \forall \epsilon > 0 : J \left [ { y^* } \right ] < \mu + \epsilon $

By assumption of continuity of $ J $:


 * $ \displaystyle \forall \epsilon > 0 : \exists \delta \left ( { \epsilon } \right ) > 0 : \left ( { \left \vert y - y^* \right \vert < \delta } \right ) \implies \left ( { \left \vert J \left [ { y } \right ] - J \left [ { y^* } \right ] \right \vert < \epsilon } \right ) $

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


 * $ \displaystyle \left \vert \eta_n - y^* \right \vert < \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 \left [ { y_n } \right ] $

Hence:


 * $ \displaystyle J \left [ { y_n } \right ] < J \left [ { \eta_n } \right ] + \epsilon < \mu + 2 \epsilon $

On the other hand, by the definition of $ y_n $:


 * $ \displaystyle \mu \le J \left [ { y_n } \right ] \le J \left [ { \eta_n } \right ] $

Thus:


 * $ \mu \le J \left [ { y_n } \right ] < \mu + 2 \epsilon $

If $ J \left [ { y_n } \right ] = \mu $, then the inequality is satisfied, and $ n $ is finite.

If $ \mu < J \left [ { y_n } \right ] $, then subratct $ \mu $ from all the terms:


 * $ 0 < J \left [ { y_n } \right ] - \mu < 2 \epsilon $

Obviously:


 * $ \left ( { \epsilon > 0 } \right ) \implies \left ( { - 2 \epsilon < 0 } \right ) $

Hence:


 * $ - 2 \epsilon < J \left [ { y_n } \right ] - \mu < 2 \epsilon $

or


 * $ \left \vert J \left [ { y_n } \right ] - \mu \right \vert < 2 \epsilon $