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Given the Banach space $l_p = \left\{f\in R^n : \|f\| \leq \infty\right\}$ for $1\leq p \leq \infty$, we define the following norms

  • $\|f\|_p = \left(\displaystyle\sum_{j=0}^{\infty} |f_j|^p\,w_j\right)^{\frac{1}{p}},\quad p\geq 1,\,w_j>0,\,\forall j\in\mathbb N$
  • $\|f\|_{\infty} = \underset{j\in\mathbb N}{\sup} |f_j|$

I want to prove that the spaces $l_1$ and $l_{\infty}$ are not strictly convex.

For $l_1$, I'm trying to find two different sequences $\left\{a_n\right\}_{n=1}^{\infty}$ and $\left\{b_n\right\}_{n=1}^{\infty}$ such that

$$\displaystyle\sum_{j=0}^{\infty} |a_n| + \displaystyle\sum_{j=0}^{\infty} |b_n| = \displaystyle\sum_{j=0}^{\infty} |a_n+b_n|$$

Respectively, for $l_{\infty}$, I have to find two sequences such that

$$\underset{j\in\mathbb N}{\sup} |a_j|+ \underset{j\in\mathbb N}{\sup} |b_j| = \underset{j\in\mathbb N}{\sup} |a_j+b_j|$$

are these argument correct?

If they are, then I think that I can prove it for $l_{\infty}$. One just has to take the sequences:

$$\left\{a_n\right\}_{n=1}^{\infty}=\left\{\frac{1}{2^{n}}\right\}_{n=1}^{\infty} $$ $$\left\{b_n\right\}_{n=1}^{\infty}=\left\{\frac{1}{2n}\right\}_{n=1}^{\infty}$$

Since then we have:

$$\underset{j\in\mathbb N}{\sup} |a_j|+ \underset{j\in\mathbb N}{\sup} |b_j| = \frac{1}{2} + \frac{1}{2} = 1 = \underset{j\in\mathbb N}{\sup} |a_j+b_j|$$

synack
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    For both spaces, just take $(1,0,0,\ldots)$ and $(0,1,0,0,\ldots)$. – David Mitra May 20 '13 at 15:32
  • By the way, you need to find two non-zero elements that are not multiples of each other such that your equalities hold. – David Mitra May 20 '13 at 15:36
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    It is a good idea to draw the unit balls of $\ell^1$ and $\ell^\infty$ in dimension $2$ to get a visual idea of what is going on. – Julien May 20 '13 at 15:39
  • I like your solution, @DavidMitra, but I don't understand your second comment, why is that necessary? – synack May 20 '13 at 15:48
  • For any space, e.g., $\Vert x\Vert +\Vert 2x\Vert=\Vert x+2x\Vert$. – David Mitra May 20 '13 at 15:51
  • So, $a_n$ and $b_n$ not only have to be different but also cannot be multiples? – synack May 20 '13 at 15:53
  • The vectors can't be multiples of each other (particular coordinates may be). For instance, $(1,0,\ldots)$ and $(2,0,\ldots)$ do not show that $\ell_1$ is not strictly convex; but the vectors $(1,1,0,\ldots)$ and $(2,1,0,\ldots)$ do. – David Mitra May 20 '13 at 16:00
  • Mmmh.. I'm not sure that I understand. Anyway, the vectors you proposed in the first comment don't work for the $l_{\infty}$. – synack May 20 '13 at 16:03
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    Ah, sorry. For $\ell_\infty$, you could take $(1,1,0,0,\ldots)$ and $(1,0,0,\ldots)$. As for "not being multiples", if this were allowed in the definition, then no space would be strictly convex. – David Mitra May 20 '13 at 17:06

2 Answers2

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I like the geometric interpretation of "non-strictly convex": the unit sphere $\{x:\|x\|=1\}$ contains a nondegenerate line segment (line segment is degenerate if it has only one point). If one keeps this in mind and follows julien's suggestion

It is a good idea to draw the unit balls of $\ell_1$ and $\ell_\infty$ in dimension $2$ to get a visual idea of what is going on

then the answer comes out naturally: the line segment with endpoints $e_1$ and $e_2$ (or any two vectors of the standard basis) lies on the unit sphere of $\ell_1$; also, the line segment with endpoints $e_1$ and $e_1+e_2$ lies on the unit sphere of $\ell_\infty$. This is where the examples given by David Mitra come from.

ˈjuː.zɚ79365
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I think

$$\displaystyle\sum_{j=0}^{\infty} |a_n| + \displaystyle\sum_{j=0}^{\infty} |b_n| = \displaystyle\sum_{j=0}^{\infty} |a_n+b_n|$$

is not correct for all situations.

Consider the following $$\left\{a_n\right\}_{n=0}^{\infty}=\left\{-\frac{1}{2^{n}}\right\}_{n=0}^{\infty} $$ $$\left\{b_n\right\}_{n=0}^{\infty}=\left\{\frac{1}{2^{n+1}}\right\}_{n=0}^{\infty}$$

$$\sum\limits_{j=0}^{\infty}|-\frac{1}{2^n}|+\sum\limits_{j=0}^{\infty}|\frac{1}{2^{n+1}}| > \sum\limits_{j=0}^{\infty}|-\frac{1}{2^n}+\frac{1}{2^{n+1}}|=\sum\limits_{j=0}^{\infty}|-\frac{1}{2^{n+1}}|=\sum\limits_{j=0}^{\infty}|\frac{1}{2^{n+1}}|$$

Jimmy Wang
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    I don't understand what you mean. To prove that they are not strictly convex I just need to find a counterexample for which the equality does hold, right? – synack May 20 '13 at 16:20
  • If you really check the example, you can find left hand is larger than right. – Jimmy Wang May 20 '13 at 16:46
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    The problem is to find an example of two appropriate vectors for which the equality holds. It is not claimed that the equality holds for all pairs of vectors. – David Mitra May 20 '13 at 16:55
  • In that sense, "not strictly convex" should be explained more clear. Like in real function, $f''$ is positive more strict than non-negative? – Jimmy Wang May 20 '13 at 17:00