I have trouble understanding the following statement (From Gelfland's Calculus of Variations book):

If $\phi[h]$ is a linear functional and if

$$\frac{\phi[h]}{\left|\left|h\right|\right|}\rightarrow 0,$$

as $||h||\rightarrow0$, then $\phi[h]=0$ for all $h$. In fact, suppose $\phi[h_0]\neq0$ for some $h_0\neq0$. Then, setting


we see that $||h_n||\rightarrow 0$ as $n\rightarrow \infty,$ but


contrary to hypothesis.

In the above $h$ belongs to some normed linear (function) space $\mathcal{F}$, $||\cdot||$ is the norm function in that space and $\phi[\cdot]$ is a linear functional.

My question is about the part:

$\displaystyle\frac{\phi[h]}{\left|\left|h\right|\right|}\rightarrow 0,\;\;\;$ as $||h||\rightarrow0$, then $\phi[h]=0$ for all $h$.

I don't understand why does the $\phi[h]$ need to equal $0$? Isn't it enough that $\phi[h]\rightarrow 0$ faster than $||h||$?

Thank you for your help! Please let me know if you need more info.

Here is a picture of the part in my book. I have highlighted the area, where I have problems

enter image description here

  • $\begingroup$ Could you explain a little bit? I mean you quoted a proof for this theorem. Have you understand it? $\endgroup$ – tired Jan 29 '15 at 11:56
  • $\begingroup$ Hi @tired this is part of a theorem in my book. No I have not understood the theorem because I got stuck with this one. Do you want more details from the book? The theorem's claim is: "The differential of a differentiable functional is unique" I will add a picture of the theorem here asap :) $\endgroup$ – jjepsuomi Jan 29 '15 at 11:58

You are given a space ${\cal F}$ and a linear functional $\phi:\>{\cal F}\to{\mathbb C}$ that you are trying to better understand. The only thing you are told is that $$\lim_{h\to0}{\phi(h)\over\|h\|}=0\ .$$ The statement in question says that in such a case one necessarily has $\phi(x)\equiv0$ on ${\cal F}$.

For a proof consider an arbitrary $x\in{\cal F}$, $x\ne0$. Then $$\phi(x)=\|x\|{\phi(\lambda x)\over \|\lambda x\|}\qquad\forall \lambda>0$$ and therefore $$\phi(x)=\|x\|\>\lim_{\lambda\to 0}{\phi(\lambda x)\over \|\lambda x\|}=0\ .$$

  • $\begingroup$ Excellent, that one did it! =) Thank you! $\endgroup$ – jjepsuomi Jan 29 '15 at 12:55
  • $\begingroup$ I think my problem was mostly with the notation. $\endgroup$ – jjepsuomi Jan 29 '15 at 13:04

The notation isn't the best. I would write the theorem like this: if $\phi$ is a linear functional on $\mathcal{F}$ such that, for any unit vector $u\in\mathcal{F}$, $\frac{\phi[tu]}{|t|}\to0$ as $|t|\to0$, then $\phi=0$ identically. But written like this it becomes obvious since $\frac{\phi[tu]}{t}=\phi[u]$, so the condition pretty much says that $\phi[u]=0$ for all unit vectors $u\in\mathcal{F}$, which of course implies $\phi=0$.

  • $\begingroup$ Thank you for your help! =) $\endgroup$ – jjepsuomi Jan 29 '15 at 12:14
  • $\begingroup$ But, I still did not get it 100% sry :) Because the theorem didn't state any assumptions that $h$ should be a unit vector. Or does it follow that $h$ needs to be a unit vector? $\endgroup$ – jjepsuomi Jan 29 '15 at 12:36
  • 1
    $\begingroup$ Note that $h=\|h\|u$ for a unit vector $u$ - I have replaced $\|h\|$ by $t$ so that we have a fixed unit vector. $\endgroup$ – Jason Jan 29 '15 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.