Proof of non-strictly convexity of $l_1$ and $l_{\infty}$

Question

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:

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

Answer 1

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

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

Answer 2

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.