I do not understand exactly why $1 <= K(x - a)$ implied that $x = a$

$$ |f(x_\varepsilon)| \le K\int_{a}^{x_\varepsilon}|f(t)|dt \le K(x_\varepsilon-a)\varepsilon $$ How do you derive the left-hand-side inequality?!

I edited the post, refresh.

How do you know that the maximum of function $g$ will be epsilon?! Instead of epsilon, there should be the supremum of the function $g$ on that interval (i.e. absolute value of $f$)

This follows from the definition of $x_\varepsilon$.

Can you please exemplify more on the set of $x_{\epsilon}$?! I still can not figure out the meaning of this set. (pretty sure this comes from continuity of $f$, but I would be really appreciated if you could provide a little more details if possible)

By definition of $x_\varepsilon$, there holds that $|f(t)|\le\varepsilon$ for all $t\le x_\varepsilon$ (this has nothing to do with the continuity of $f$, it is by definition of the infimum).

So it means that $x_{\epsilon}$ is the smallest ("infimum") number for which $x_{\epsilon} > a$ and $|f(x_{\epsilon})| >= \epsilon$? Meaning that for all numbers $y$ less than $x_{\epsilon}$, we will know that $|f(y)|$ is less than $\epsilon$?

No, $x_\varepsilon$ is the largest lower bound for all $x>a$ with $|f(x)|\ge \varepsilon$. This directly implies, that $|f(t)|\le \varepsilon$ for all $t<x_\varepsilon.$ Moreover, by continuity of $f$ there holds $|f(x_\varepsilon)|\ge \varepsilon$.

Oh, now I see the point. But now arises another question: how continuity of $f$ will guarantee that for any small values of $\epsilon$, you can find that the function values near the neighbourhood will be larger than $\epsilon$? I guess there should be another point $y_0$ if we are dealing with continuity

Moreover, you obtained that $x_{\epsilon}$ is larger than one quantity. How can you take the limits for each $\epsilon$ where you do not have control over the choices of $\epsilon$

I do not claim this. The assumption $f\neq 0$ implies that $x_\varepsilon$ is-well defined. Moreover, continuity of $f$ guarantees that $x_\varepsilon \to a$ as $\varepsilon\to 0$. On the other hand, the calculation shows, that for any $\varepsilon>0$ we have that $x_\varepsilon >\frac{1}{K}+a$ (note that this bound is independent of $\varepsilon$!), which is a contradiction to $x_\varepsilon\to a$. Hence, $f=0$.

I agree with the well-definition of $x_{\epsilon}$. But, how do you get that $x_{\epsilon}$ approaches to $a$ whenever $\epsilon$ approaches to zero?!

This is because $f(a)=0$ and $f$ is continuous.

Honestly, the last part really lacks the mathematical rigor: how $f(a) = 0$ and continuity of $f$ directly implies that limit of $x_{\epsilon}$ approaches to $a$ whenever $\epsilon$ approaches to zero: This really requires mathematical formalism, intuitively this seems okay though (however, I am still confused on how you choose these $\epsilon$ values, as this is not correct to assume that you have the control of the choices of these $\epsilon$ values.

I correct the post, wait some minutes

thank you so much!

I am pretty sure that your logic is correc

t

it is just more mathematical rigor that is lacking

No problem, you were right. There was some flaw in the argument, we need a preliminary step