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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 $\overline{x}_\alpha \to \overline{x}$.

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