# Prove that $|f(y) - f(x)| \leq |y-x| \left ( \sup_{z \in \alpha} |f'(z)| + \epsilon \right )$

Let $$X$$ be a subset of $$\mathbb C$$ and let $$\alpha: [0,1] \to X$$ be a curve in $$X$$.

Question: Given $$\epsilon > 0$$ and $$f$$ a function with continuous derivative in $$X$$, prove that for each $$x \in \alpha$$, exists a neighborhood $$U_x$$ of $$x$$ such that $$|f(y) - f(x)| \leq |y-x| \left ( |f'|_\alpha + \epsilon \right ), \, \forall y \in U_x$$

where $$|f'|_\alpha = \sup_{z \in \alpha} |f'(z)|$$.

My attempt: For each $$y$$, define $$\gamma_y (t) = tx + (1-t)y, \, t \in [0,1]$$. Then, $$|f(x) - f(y)| \leq \left | \int_{\gamma_y} f'(\zeta)\, d\zeta \right | \leq |f'|_{\gamma_y} \, L({\gamma_y})$$

Since $$L(\gamma_y) = |x-y|$$, I just need that $$|f'|_{\gamma_y} \leq |f'|_\alpha + \epsilon$$. If I define

$$U_x = \left \{ y \in X: |f'|_{\gamma_y} \leq |f'|_\alpha + \epsilon \right \}$$

we have that $$x\in U_x$$, hence $$U_x \neq \emptyset$$. However, I'm not so sure that this $$U_x$$ is an open set.

Help?

For a given $x \in \alpha$, lets consider the function $$h(y)=|\frac{f(x)-f(y)}{x-y}|$$ Since $f(x)$ is differentiable, we know that $h(y) \to |f'(x)|$ as $y \to x$. Given now $\epsilon > 0$, we can find find $\delta >0$ s.t. $|x-y|<\delta$ implies: $$h(y) \leq h(x)+\epsilon$$ In other words, we found $U_x$ open so that we can estime for all $y \in U_x \cap X$: $$|\frac{f(x)-f(y)}{x-y}|\leq|f'(x)|+\epsilon \leq |f'|_{\alpha} +\epsilon$$ Now you can multiply with $|x-y|$ to get the result. However, this is a property of differentiable functions rather than having to do sth with curves in a compact space. Can you please double-check that you included every information/detail?