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.

  • $\begingroup$ Your reasoning is correct. The fact that $h$ is a function and takes some other argument does not matter, since, to use your own words, you are viewing $h$ as a point in $\mathbb{F}$. $\endgroup$ – angryavian Jun 21 '17 at 22:55
  • $\begingroup$ @angryavian: Would a possible functional be $L [h(x)] = x $, so that it depends on $x $? or would this not be a functional, as the RHS is not a function of $h$, but rather $h $'s argument (making this just a composite function on $\mathbb {R}$)? $\endgroup$ – infinitylord Jun 21 '17 at 23:02
  • $\begingroup$ The functional you have written is a constant functional, since it does not depend on $h$. Moreover $x$ is fixed, so it "depends on $x$," but only through $L$; it is the same way that your $\epsilon$ and $\delta$ would depend on $L$ regardless. A more interesting functional would be, for a fixed $x$, the functional $L[h] = h(x)$. But again, things will depend on $x$ only through $L$. $\endgroup$ – angryavian Jun 21 '17 at 23:53
  • $\begingroup$ @angryavian: okay, thank you. I hadn't realized that $x $ was considered fixed when we considered $L[h(x)] $. I believe this means that I was correct in assuming that $L[h(x)] $ is single valued (meaning that it always equals the same value, the same way that $h(x) $ does for a fixed $x \in \mathbb{R} $)? Originally I was concerned that we could plug in, say, $L[h(x)] $ and $L[h(y)] $ for $x \ne y $ and get potentially different values at $h$. But I suppose that would contradict the fact that the input is a function. $\endgroup$ – infinitylord Jun 22 '17 at 0:09

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 $f$.

  • $\begingroup$ Is this to say that a linear functional is continuous iff it's norm is bounded by the norm of its input? And wouldn't the norm of $L$ just be the standard norm on $\mathbb{R} $ Since it is mapping to a real number? The norm on the function (input) would still need to be defined as I see it (perhaps as the supremum on an interval) $\endgroup$ – infinitylord Jun 22 '17 at 0:20
  • $\begingroup$ That's right, you just take the regular norm in $\mathbb{R}$. And yes, the norm defined by taken the supremum (over the whole space) makes $\mathbb{F}$ a Banach space. $\endgroup$ – matboy Jun 22 '17 at 0:30
  • $\begingroup$ Edit: It's a Banach space under certain conditions, for example is the functions are defined on a compact space. There seem to be quite a few conditions, but they're general ones $\endgroup$ – matboy Jun 22 '17 at 0:39

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