For each of the following decide whether the suggested formula defines a norm on the indicated space. You may assume that $||f||_1=\int_0^1 |f(t)| dt$ does give a norm on the space of all continuous functions on the interval $[0,1]$

$$V_a=\mathbb{R}^2 \ \ \ \ \ \ \ \ ||(x,y)||_a=|x+y|$$

$$V_b=\mathbb{R}^2 \ \ \ \ \ \ \ \ ||(x,y)||_b=max(|x|,|y|)$$

$$V_c=\mathbb{R}^2 \ \ \ \ \ \ \ \ ||(x,y)||_c=\int_0^1|x+yt|dt$$

$$V_d \ \ \ \ \ \ \ ||f||_d=\int_0^1|f'(t)|dt$$ where $V_d$ is the space of all differentiable functions $f$ on $[0,1]$ with $f'$ continuous.

The first 2 ($V_a$ and $V_b$) I know are norm spaces. The 3rd one $V_c$, the $t$ variable makes me nervous. Not sure how to work around it. But I ended up getting that it does satisfy all norm space properties. $V_d$ is not a norm space because if $f$ is constant, it would fail separates points axiom. Can anyone concur? Thank you.

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    $\begingroup$ I guess there is an error in the fourth one. $\endgroup$ – Alexander Grothendieck May 20 '14 at 22:18
  • $\begingroup$ @E.T. Oops sorry I edited. $\endgroup$ – User69127 May 20 '14 at 22:22
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    $\begingroup$ The fist one is not a norm. Note that $||(x,-x)||_a=|x+(-x)|=0,$ for any $x\in \mathbb{R}.$ $\endgroup$ – mfl May 20 '14 at 22:25
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    $\begingroup$ $V_a$ is not a normed space, the line $x+y=0$ lies in the null-space of that "norm". $\endgroup$ – TZakrevskiy May 20 '14 at 22:26

$V_c$ (which is apparantly the only one which is a problem for you) is a normed space, notice that if $$\int_0^1|x+ty|dt=0$$ then it follows (since you may assume the standard norm on $C^0[0,1]$) that $|x+ty|=0$ for all $t\in[0,1]$, hence $(x,y)=(0,0)$. Can you do the triangle inequality?

  • $\begingroup$ I think so. I I just assumed for some $||(a,b)||$ $\int_0^1|x+yt+a+bt|dt \le \int_0^1|x+yt|dt+\int_0^1|a+bt|dt$ $\endgroup$ – User69127 May 20 '14 at 22:40
  • $\begingroup$ very true, this suffices. $\endgroup$ – Alexander Grothendieck May 20 '14 at 22:41
  • $\begingroup$ What about the error on the 4th one? $\endgroup$ – User69127 May 20 '14 at 22:44
  • $\begingroup$ well its just not clear what you mean by that norm, I mean is $f$ fixed? $\endgroup$ – Alexander Grothendieck May 20 '14 at 22:56
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    $\begingroup$ @User69127: yeah it confused me for a minute or so. Yes it is! By the way your space $V_d$ is usually denoted $C^1[0,1]$. $\endgroup$ – Alexander Grothendieck May 20 '14 at 23:10

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