min max problem with eigenvalues

Question

How can I find $$\text{s} = \inf_{a\in R}\left\{ \max_{λ\in[λ_n,λ_m]}|1-αλ|\right\}$$

I guess that λ_n and λ_m are eigenvalues of some descent method but there are no more clues in the question

Answer 1

For a fixed $\alpha$, $|1-\alpha\lambda|$ is an absolute value function centered at $1/\alpha$ with slope $\alpha$.

Thus, the maximum over $[\lambda_n,\lambda_m]$ will occur at whichever of $\lambda_n,\lambda_m$ is further from $1/\alpha$, where it will attain a value of $|1-\alpha \lambda_m|$ or $|1-\alpha\lambda_n|$.

Thus, we have, $$ s = \inf_{\alpha} \max\{ |1-\alpha\lambda_m|, |1-\alpha\lambda_n| \} $$

Suppose $\lambda_n < 0 <\lambda_m$. Then $\alpha=0$ and $s=1$.

Otherwise, we need $\alpha \in [\lambda_n,\lambda_m]$.

Now, $|1-\alpha \lambda_m|$ is an absolute value function centered at $1/\lambda_m$ with slope $\lambda_m$. Likewise, $|1-\alpha\lambda_n$ an absolute value function centered at $1/\lambda_n$ with slope $\lambda_n$.

If you think about what these plots look like its clear that the infimum must occur when $|1-\alpha \lambda_m| = |1-\alpha \lambda_n|$. Thus, $\alpha = 2/(\lambda_n+\lambda_m)$ and $$ s=1 - 2 \lambda_n / (\lambda_n+\lambda_m) = -(1-2 \lambda_m / (\lambda_n+\lambda_m)) $$