# What is a conceptual understanding of what it means for evaluation functionals to be continuous?

I was trying to understand what it means for evaluation functionals to be continuous. First recall that evaluation functionals are maps $e_x:\mathcal{H} \rightarrow \mathbb{R}$ such that $e_x(f) = f(x)$ (i.e. evaluate functions $f \in \mathcal{H}$). If we just apply the definition of continuous we get:

$$lim_{f \rightarrow f'} e_x(f) = e_x(f')$$

or equivalently:

$$lim_{f \rightarrow f'} e_x(f) = f'(x) = f'_x$$

we can go further and unpack the definition using epsilon delta:

$$\forall x \in \mathcal{X},\forall \epsilon>0, \exists\delta(x,\epsilon)>0, if \| f - f' \|_{\mathcal{H} } <\delta(x,\epsilon) \implies \|f(x) - f'(x) \|_{R} < \epsilon$$

which I guess means that given an epsilon we can always find some constant $\delta(x,\epsilon)$ (that depends on the point $x \in \mathcal{X}$) for which if the functions $f$,$f'$ are close enough according to $\delta(x,\epsilon)$ then the evaluation at any (given) point $x \in \mathcal{X}$ is close according to the prescribed accuracy $\epsilon$. Which this is my attempt to make sense of the definition using words. If anything I said is wrong or omitting something important that can shed some conceptual/intuitive understanding, feel free to jump in.

But for me it seems sort of a weird definition/requirement because it says that if we have two functions that are really close (according to $\mathcal{H}$, maybe a Hilbert space say) then we have that when we evaluate them at any/all points, then they are as close as we want. Which seems really weird, because of course if we have two functions that are the same or getting very close according to the metric endowed by $\mathcal{H}$, then of course they will output the same number when evaluated at any $x$/. It seems nearly tautological.

1) Can someone clarify why its not tautological? What function spaces don't have the property that when two functions are really close they don't behave the same when evaluated at points?

I didn't give up there. I had a hunch that my misconception was actually due to the definition of a limit. I understand that in normal calculus at the exact point $f'$ the function might have some different value (or even undefined), but it might still have a limit. In this goal I tried to visualize what might be going on (with a picture that might be wrong or not general enough): (I understand this picture is probably wrong in many ways but I had to try to understand this in someway. Some of the issues that it might have is that I have basically pictured functions as just points in a real line, which seems questionable. Even if a norm $\| \cdot \|_{\mathcal{H}}$ exists, it might not be possible to order the points in a line the way I did...anyway, I didn't have better tools to understand this so I went ahead and drew the picture).

I guess from this picture and this discussion my main question in this section is, sure I can define a "discontinuity" in this picture but I can't understand what that means. What does it mean to have a discontinuity in a evaluation functional? Does it mean that for that $\forall x \in \mathcal{X}$, that some specific function $f'$ doesn't evaluate what we would expect if we followed the limit of functions? Now that I've said this, I don't quite even understand what a limit of functions means. i.e. if $f \rightarrow f'$ means at a conceptual level. If we had a sequence of functions $\{ f_n \}_{n}$ then maybe I would understand better what this means limit means since having a sequence seems to define an ordering better and its much more intuitive to follow a list with orderings that some abstract functions $f$, $f'$. Maybe there isn't a sequence because it doesn't matter from which "direction" (or order) we approach $f'$. If the distance between the two functions is small enough (with respect to some quality/metric of the function according to $\mathcal{H}$ then where you approach $f'$ doesn't really matter?)

But basically it seems this limit is charged with a lot of weight (and meaning) in so many dimensions. It seems I need to understand:

1. what a discontinuity in a evaluation functional means,
2. what a limit of functions means (or even what a sequence of functions is or why maybe such a sequence doesn't even exist).

Furthermore, I remember hearing my labmate/friend say "In particular this implies that convergence with respect to the norm in $\mathcal{H}$ implies point-wise convergence". Which don't think I fully understand or appreciate what it means.

• Elements of Hilbert spaces usually are generalized functions and do not have clearly defined function values. -- See pointwise convergence vs. convergence in a function space norm. – LutzL Sep 25 '16 at 17:47
• If you take $\mathcal H = L^2([0,1])$, then as @LutzL points out, evaluation isn't well-defined. But I suggest trying to prove or disprove the following statement: for a sequence of continuous functions $f_n$ on $[0,1]$ that converge to $f$ in $L^2([0,1])$, one has $e_x(f_n) \to e_x(f)$ for all $x\in[0,1]$. – user360874 Sep 25 '16 at 17:52
• @LutzL it was easy to find what pointwise convergence means online. However, I can't quite find the second term i.e. "convergence in a normed function space". Do you happen to have any link you recommend to make sure I am reading the right thing? Is uniform convergence the same thing as the second term you refer to? – Charlie Parker Sep 25 '16 at 19:11
• @user37208 I was trying to do the exercise you left me. However, I came across a definitional issue I don't understand. In the context of the problem you left me, what does $f_n$ converges to $f$ mean? In particular what metric to compare $f_n$ and $f$ did you have in mind? Does it mean $\forall \epsilon > 0, \exists N > 0$ if $n > N$, then $\| f_n - f \|_{\mathcal{H}} < \epsilon$? In particular what did you have in mind for defining $\| f_n - f \|_{\mathcal{H}}$? For some reason I have a hunch the answer is "false" depending on that metric is defined... – Charlie Parker Sep 25 '16 at 21:46
• @CharlieParker I mean convergence in the $L^2$ norm, i.e. $\int_0^1 |f_n(x) - f(x)|^2 dx \to 0$ as $n\to\infty$. – user360874 Sep 25 '16 at 22:05