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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

$$h_n=\frac{h_0}{n},\;\;\;\;\lambda=\frac{\phi[h_0]}{||h_0||},$$

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

$$\lim_{n\rightarrow\infty}\frac{\phi[h_n]}{||h_n||}=\lim_{n\rightarrow\infty}\frac{n\phi[h_0]}{n||h_0||}=\lambda\neq0,$$

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

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  • $\begingroup$ 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
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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 $$\phi(x)=\|x\|\>\lim_{\lambda\to 0}{\phi(\lambda x)\over \|\lambda x\|}=0\ .$$

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  • $\begingroup$ 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
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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 $\phi=0$.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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