0
$\begingroup$

Consider a positive semi-definite function $f(x) \in \mathcal{L}^2\left( \mathbb{R} \right)$ of class $\textbf{C}^0$. I am interested in knowing the values of $x$ beyond which this function is uniformly below some value $\overline{f}$.

$$ \overline{x} \triangleq \min \left\{ x \, \Big \vert \, f\left( r\right) \leq \overline{f}, \,\, \forall r > x \right\} $$

We will call $\overline{x}$ to be the cut-off point of $f(x)$, and we will assume that $\overline{x}$ exists.

Let $\overline{x}_\alpha$ represent the cut-off point of the function $\alpha f(x)$, where $\alpha$ is a positive real number, then I would like to prove that $$ \lim_{\alpha \to 1} \overline{x}_\alpha =\overline{x} $$

Here too, we will assume that $\overline{x}_\alpha$ exists.

My tentative approach for the proof

In the case of an affine function $f(s) = mx + c$, then for any given $\alpha$, we can find an exact value of $\overline{x}_\alpha$, which is $$ \overline{x}_\alpha = \left(\frac{1-\alpha}{\alpha}\right)\frac{\overline{f}}{m}$$

In the affine case, $\lim_{\alpha \to 1} \overline{x}_\alpha = \overline{x}$.

For other functions, we can find an approximate value for $\overline{x}_\alpha$ using an affine approximation the function $\alpha f(x)$
$$ \hat{\overline{x}}_\alpha = \left(\frac{1 - \alpha}{\alpha}\right) \frac{\overline{f}}{f'(\overline{x})} $$

In case $f'(x)$ does not exist, then we could replace it by $f'(\overline{x}^-)$ or $f'(\overline{x}^+)$ depending on whether $\alpha < 1$ or $\alpha > 1$.

Depending on the value of $\alpha$ and the natuee of $f(x)$, we know that $\hat{\overline{x}}_\alpha \approx \overline{x}$.

However, we know that $\lim_{\alpha \to 1} \hat{\overline{x}}_\alpha = \overline{x}$. Does this also mean that $\lim_{\alpha \to 1} \overline{x}_\alpha = \overline(x)$? If so, how can we demonstrate this?

I think one way to do this could be to show that $\alpha \to 1$, implies that $\hat{\overline{x}}_\alpha \to \overline{x}_\alpha$. This would in turn imply that $\overline{x}_\alpha \to \overline{x}$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.