# Continuity of a solution of the equation $f(x) - \overline{f} = 0$.

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