Ritz Method implies Not Worse Approximation with Increased Number of Functions

Theorem
Consider Ritz method.

Let $ \eta_n = \boldsymbol \alpha \boldsymbol \phi $, and $ J \left [ { \eta_n } \right ] = \mu_n $.

Then $ \mu_n \ge \mu_{ n + 1 }$

Proof
Denote $ \eta_{ n + 1 } = \eta_n + \alpha_{ n + 1 } \phi_{ n + 1 } $.

Clearly, for $ \alpha_{ n + 1 } = 0 $, $ \eta_n = \eta_{ n + 1 } $.

Suppose, $ J \left [ { \eta_n } \right ] $ has been minimised $ \boldsymbol \alpha$.

If:


 * $ \exists \boldsymbol \alpha \in \R^n : \not \exists \alpha_{ n + 1 } \ne 0 : J \left [ { \eta_n } \right ] > J \left [ { \eta_{ n + 1 } } \right ] $

then $ J \left [ { \eta_{ n + 1} } \right ] $ is minimised for previously determined $ \boldsymbol \alpha $ and $ \alpha_{ n + 1 } = 0 $:


 * $ J \left [ { \eta_{ n + 1} } \right ] = J \left [ { \eta_{ n } } \right ] $

where $ \eta_{ n + 1 } = \eta_n $

Suppose:


 * $ \exists \boldsymbol \alpha \in \R^n : \exists \alpha_{ n + 1 } \ne 0 : J \left [ { \eta_n } \right ] > J \left [ { \eta_{ n + 1 } } \right ] $

Then, $ \eta_{ n + 1 } \ne \eta_n $, because their respective $ \boldsymbol \alpha $ differ by at least one value, $ \alpha_{ n + 1 } $.

Hence, for this supposition $ J \left [ { \eta_{ n + 1} } \right ] > J \left [ { \eta_{ n } } \right ] $.

Finally, both cases together imply that:


 * $ J \left [ { \eta_{ n + 1} } \right ] \ge J \left [ { \eta_{ n } } \right ] $

or, equivalently:


 * $ \mu_n \ge \mu_{ n + 1 } $