Does the following limit hold uniformly for all $f$ in a function space $X$?

Question

Suppose I have the following for all $\epsilon > 0$ $$F(f,h) \leq C_0\epsilon + C_1(\epsilon)\sqrt{h}\quad\text{for all $f \in X$}$$ where $F:X\times \mathbb{R} \to \mathbb{R}$ and $X$ is some subset of a Banach space, and $C_0$ is a constant and $C_1(\epsilon)$ is a constant depending on $\epsilon.$

My questions:

1. Does it follow that $\lim_{h \to 0}F(f,h) = 0$?

   I think this is true since obviously $\sqrt h \to 0$ and the above holds for every $\epsilon > 0$.

2. Does it follow that $lim_{h \to 0}F(f,h) = 0$ uniformly for $f \in X$?

   Also I think this is true since the RHS of the inequality above is independent of $f$ it should be uniform.

Answer 1

Assuming that $F(f,h) \geqslant 0$ - otherwise it could be negative of arbitrarily large modulus - we indeed have

$$\lim_{h\to 0} F(f,h) = 0\tag{1}$$

uniformly on $X$.

As you note, the right hand side does not depend on $f$, so any limiting behaviour can be dominated independently of $f$ if at all.

To see $(1)$, fix an arbitrary $\eta > 0$. For that $\eta$, choose $\epsilon > 0$ so that $C_0\epsilon < \eta/2$. Then for

$$\sqrt{h} < \frac{\eta}{2C_1(\epsilon)}$$

we have

$$C_0\epsilon + C_1(\epsilon)\sqrt{h} < \eta.$$