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I recently took real analysis at my school, and enjoyed it thoroughly. I decided to use some of my Summer to study variational calculus, and wanted some verification of my work (or correct work if I'm wrong) for this example.

Let the functional $\phi$ be defined by: $$\phi[h] = h(x_0) \space\text { for each function } \space h(x) \in \mathbb{F}(a,b) $$ Where $\mathbb {F}(a,b)$ is the set of continuous functions on the interval $(a,b) $ and $x_0 \in (a,b) $

Prove that $\phi [h] $ is a continuous linear functional on this function space.

The linearity part was trivial, I just want verification in prove continuity.

$\phi $ is continuous at $h(x)$ provided that $\forall \epsilon >0, \exists \delta > 0 \text { s.t. }$ $$||h(x)-g(x)||<\delta \implies |\phi[h]-\phi [g]|<\epsilon$$ Where the norm on $\mathbb{F}(a,b) $ is defined as: $$||h(x)|| = \max_{a \le x \le b} |h(x)|$$.

My thought was that if we let $\delta = \frac{||h(x)-g(x)||}{|h(x_0) - g (x_0)|}\epsilon $, Then we are done.

Is this an acceptable approach? I thought it would work, since $\delta $ is only a function of $\epsilon $ for any given $g $, but if we were to adapt it to a general case, this argument could prove (incorrectly) that anything is continuous, so I'm very wary of this style.

Could someone explain:

1) where my argument fails if it is incorrect, and provide a correct one

2) where this style argument would fail for a general functional if it is correct

Jonas Dahlbæk
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infinitylord
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  • if you work with the norm ||h|| = max|h(x)|, you should make sure to operate on the the closed set [a,b], otherwise the maximum does not necesserily exist for cont. functions h. – Simonsays Jun 21 '17 at 13:46
  • @Simonsays: I was just using the definition stated in the book, but I see your point. For argument's sake, let's assume the interval is closed (or use the supremum instead of maximum). – infinitylord Jun 21 '17 at 13:50
  • ok good to know: than I have something for you :) – Simonsays Jun 21 '17 at 13:52
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    For continuity, there has to exist a $\delta$ which is independent of the function $g$. You cannot choose a $g$ and then a $\delta$ that works with it specifically. – Amitai Yuval Jun 21 '17 at 13:57
  • @AmitaiYuval: I agree, but the way I see it, this will be satisfied regardless of the choice in $g $ so long as $g (x_0) \ne h (x_0) $ – infinitylord Jun 21 '17 at 14:17
  • The minute you write $\delta=\mathrm{expression}$, and the expression on the right hand side has $g$ in it, it means you are doing something wrong. – Amitai Yuval Jun 21 '17 at 14:19
  • Your $\delta$ is allowed to depend on $h$ and $\epsilon$. Nothing more than that. – Amitai Yuval Jun 21 '17 at 14:20
  • since it its an linear functional, just compute the operator norm and continuity follows. You dont need that $\epsilon$ $\delta$ definition in that case – Simonsays Jun 21 '17 at 14:23
  • @Simonsays Proving continuity in this case is actually even easier than your approach. However, this is not the question at hand. I believe OP is asking about correctness of their $\epsilon$ - $\delta$ argument. – Amitai Yuval Jun 21 '17 at 15:12
  • @AmitaiYuval you are right for sure. But since OP is working on functional analytic problems, he should get used to operator norms as soon as possible – Simonsays Jun 21 '17 at 15:25
  • Indeed, I am glad to see the other approach (albeit alien to me), but I am curious about how to approach this with $\epsilon-\delta $, as that was the definition supplied to me thus far in the book. I will accept that my approach is incorrect for the same hesitation that I had in the OP (although I'm admittedly still struggling to see the the justification in this specific case) – infinitylord Jun 21 '17 at 15:25

2 Answers2

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So, here is an $\epsilon$ - $\delta$ proof.

First, as Simonsays suggests, we should work with continuous functions on a closed interval, in order for the norm to be well defined. Alternatively, we should consider only bounded functions. This way or another, let $h\in \mathbb{F}(a,b)$, and let $\epsilon>0$. Take $\delta=\epsilon$. For every $g\in\mathbb{F}(a,b)$ satisfying $\|g-h\|<\delta$, we have$$|\phi(g)-\phi(h)|=|g(x_0)-h(x_0)|\leq\max_{x\in[a,b]}|g(x)-h(x)|=\|g-h\|<\delta=\epsilon.$$ The major difference between this argument and the one in the question, is that here, $\delta$ does not depend on $g$.

Amitai Yuval
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  • Thank you for the proof! I have a question about continuity of functionals still. In Real Analysis, say we're trying to show $f(x) $ is continuous at $x_0$. Then we can choose a $\delta $ which depends on $\epsilon$ and $x_0$ (but not $x$) without issue. In this problem, it seems that $g $ is playing the role of $x $ in R.A., and $h $ the role of $x_0$. So would it, in general, be acceptable for our choice of $\delta$ to depend on $h $ if we're showing a functional is continuous at $h $? – infinitylord Jun 21 '17 at 20:00
  • @infinitylord Yes, $\delta$ may depend on $h$. – Amitai Yuval Jun 22 '17 at 00:20
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I assume now that you work on compact $[a,b]$ and you can use the following: a linear functional $\phi$ is continuous if and only if its operator norm is bounded, which means $ ||\phi|| := sup_{||h||=1,h \in \mathbb{F} }|\phi(h)| < \infty$. So if a functional is continuous, it is already Lipschitz continuous, with Lipschitz constant $||\phi||$. In that specific application $||\phi||$ is clearly bounded, since $|\phi(h)|=|h(x_0)| \leq max_{x \in [a,b]}|h(x)| = 1$ for $x_0 \in [a,b] $ and $||h||=1$.

Simonsays
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