Ritz Method implies Not Worse Approximation with Increased Number of Functions

Theorem
Consider Ritz method.

Let $\eta_n=\boldsymbol\alpha\boldsymbol\phi$, and $J\sqbrk{\eta_n}=\mu_n$.

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, $J\sqbrk{\eta_n}$ has been minimised $\boldsymbol\alpha$.

If:


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

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


 * $J\sqbrk{\eta_{n+1} }=J\sqbrk{\eta_n}$

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

Suppose:


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

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 $J\sqbrk{\eta_{n+1} }>J\sqbrk{\eta_n}$.

Finally, both cases together imply that:


 * $J\sqbrk{\eta_{n+1} }\ge J\sqbrk{\eta_n}$

or, equivalently:


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