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

Question

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?

Answer 1

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?