Continuity of functional using $\epsilon-\delta $ Let's say we have a functional $ L $ on the set of continuous functions, $\mathbb{F}$, and we're trying to show that $L$ is continuous at $h \in \mathbb {F} $ (where $h $ is a function of $x$) using $\epsilon-\delta$.
When using $\epsilon-\delta $ in Real Analysis to prove some function of $x$ is continuous at $x = x_0$, it is of standard practice to have $\delta $ be a function of $\epsilon $ and $x_0$ (function in the sense that it depends on them). 
Does this also work for Functional continuity, with $h$ playing the role of $x_0$ and any arbitrary continuous function, say $g $, that satisfies the necessary criterion of 'closeness', playing the role of $x $?
That is to say, can $\delta $ be a function of $h $ and $\epsilon $?
My intuition says yes, as the function is just a point in a function space, just as $x_0$ is a point in Euclidean space, so they are both constant on their respective spaces, but $h $ also depends on some other argument, $x$, so is not a constant (as $x_0$ is) on $\mathbb{R}$ (though $h $ may be constant to $L$, as it should always produce the same output value when taken as an input, right? Or can you have something like $L[h(x)] = x $?)
Thanks for the help, and I apologize if this isn't a well-proposed question. I'm only 5 pages into my book on Calculus of Variations, which I am currently self-studying.
 A: Yes, you can. But as you stated, since we're dealing with objects that depend of a parameter, things are a little more complicated. You're right, the idea here, to prove continuousness, is to use precisely an $\epsilon-\delta $ argument. Note that those two values ($\epsilon-\delta $) are positive ones, because we're working with distances. So, all you have to do is define a distance on your space of functions $\mathbb{F}$. Usually, the easiest way to do this is to define a norm on it, and work with the distance it induces. 
The main difficulty lies in this: when you're working on, for example, real valued functions, you lose very little information by "only" taking into account the norm of a given value. If I say the norm of $x\in\mathbb{R}$ is $1$, then it's either $x=1$ or $x=-1$. That's not the case when you're dealing with a space of functions, as there are usually, for any positive constant $\alpha\in{\mathbb{R}}$, infinitely many functions whose norm is $\alpha$. But we know how to deal with a particular, simpler kind of functions: linear ones. Then you enter into the theory of Banach spaces: under certaind conditions (namely if ${\mathbb{F}}$ is a Banach space), if $L$ is a linear funtional, then $L$ is  continuous if and only if there exists a positive constant $C$, so that, for any $f\in{\mathbb{F}}$, the norm of $L(f)$ is majored by $C$ times the norm of $f$.
