Value of $\alpha$ that minimilizes $\max_{k}|1-\alpha\lambda_k|$

Question

Let $\lambda_1,...,\lambda_n$ be strictly positive numbers. Then I am looking for the value of $\alpha$ which minimilizes the following expression. $$ \max_{k}|1-\alpha\lambda_k|. $$ It was given to me in a lecture that $\alpha=\frac{2}{\lambda_{min}+\lambda_{max}},$ but I don't understand the derivation. How is this $\alpha$ deduced?

Answer 1

Let's take a look at the function $f(x)=|1-ax|, x\in[\lambda_{min},\lambda_{max}]$.

If $a$ is chosen such that $f(x)=0$ somewhere outside $[\lambda_{min},\lambda_{max}]$, the function is a straight line over the interval. This means the highest point will be at $x=\lambda_{min}$ or at $x=\lambda_{max}$. In this case, however, the function will be lower after or before the interval. This means that we need to chose $a$ such that $f(x)=0$ somewhere inside the interval.

If $a$ is chosen such that $f(x)=0$ somewhere inside the interval, the function will "bounce", but the maximum of the function will still be at $x=\lambda_{min}$ or at $x=\lambda_{max}$. The slope after the bounce will be the negative of the slope before the bounce. This means that to minimize the maximum of the function, we need to choose $a$ such that $f(x)$ bounces perfectly in between $\lambda_{min}$ and $\lambda_{max}$, or at $x=\frac{\lambda_{min}+\lambda_{max}}{2}$.

We can use this fact to solve for $a$: $$f(\frac{\lambda_{min}+\lambda_{max}}{2})=0$$ $$|1-a\frac{\lambda_{min}+\lambda_{max}}{2}|=0$$ $$a=\frac{2}{\lambda_{min}+\lambda_{max}}$$

Because the maximum of the function is always either at $\lambda_{min}$ or $\lambda_{max}$, it doesn't matter if if $\lambda_1...\lambda_n$ is the set of all real numbers over the interval or a discrete set of numbers. The expression only depends on the maximum and minimum values of $\lambda$.

I hope this helps!