I'm reading a course and one of the exercises is about establishing whether scalar product is a well-conditioned operation. Here's their solution. They disturb each element of the vector by multiplying it by $(1+\alpha_{i})$ where $\alpha$ is really small.
And when one of the elements of the sum is equal to the negative of the rest of the elements, then the last part of the inequalities goes to $\infty$.
But how does that prove, that scalar product is poorly-conditioned? Surely I could put $f(x)=x$ and say that $f(x)<\infty$ but that doesn't mean that that calculating such a simple function is a poorly conditioned operation. How is that different here?
