0

I'm trying to calculate this limit rigorously.

Let $a,v \in \ell_1$ with $a_i >0$ for all $i$ and $\|v\|_1 := \sum_i |v_i| = 1$. Let $$ D(t) := \frac{\|a+tv\|_1 - \|a\|}{t} = \sum_i \frac{|a_i+tv_i| - |a_i|}{t} $$ is for all $t>0$. Then $$ \lim_{t \to 0^+} D(t) = \sum_i v_i. $$

Could you have a check on my attempt? Is there any cleaner way to obtain the result?

My attempt: First, we consider the map $$ \varphi:(-1, 0) \cup (0, 1) \to \mathbb R, t \mapsto D(t). $$

By convexity of the norm $\| \cdot \|_1$, we get $\varphi$ is increasing. Hence $\lim_{t\to 0^+} D(t)$ is well-defined and finite. We have $$ \begin{align*} \lim_{t\to 0^+} D(t) &= \lim_{t\to 0^+} \sum_{i=1}^m \frac{|a_i+t v_i| - |a_i|}{t_n} + \lim_{t\to 0^+} \sum_{i>m} \frac{|a_i+t v_i| - |a_i|}{t_n} \\ &= \sum_{i=1}^m v_i + \underbrace{\lim_{t\to 0^+} \sum_{i>m} \frac{|a_i+t v_i| - |a_i|}{t_n}}_{H_m}. \end{align*} $$

We have $$ \begin{align*} |H_m| &= \lim_{t \to 0^+} \left | \sum_{i>m} \frac{|a_i+t v_i| - |a_i|}{t} \right | \le \lim_{t \to 0^+} \sum_{i>m} \frac{||a_i+t v_i| - |a_i||}{t} \le \lim_{t \to 0^+} \sum_{i>m} \frac{t|v_i|}{t} = \sum_{i>m} |v_i|. \end{align*} $$

We have $$ \lim_{t\to 0^+} D(t) = \lim_m \sum_{i=1}^m v_i + \lim_m H_m. $$

On the other hand, $$ |\lim_m H_m| = \lim_m |H_m| \le \lim_m \sum_{i>m} |v_i| = 0. $$

Hence $$ \lim_{t\to 0^+} D(t) = \sum_{i=1} v_i. $$

Analyst
  • 6,351
  • (1) $t\mapsto D(t)$ need not be increasing. For example, if $v_i$'s are all negative, then $D(t)$ can actually decrease near $t = 0$. This means that the existence of the limit is not justified yet. (2) Is $t_n$ in your equation #2 a typo of $t$? (3) Again, the existence of the limit $H_m$ is not justified at this point. I would rather work with the sum $\sum_{i>m}$ inside the limit directly, before taking limit as $t\to0^+$. – Sangchul Lee Jun 04 '22 at 21:56
  • The map $a \mapsto |a|_1$ is differentiable for when all $a_i \neq 0$. – copper.hat Jun 04 '22 at 22:03
  • @SangchulLee Could you have a check on my reasoning that $\varphi$ is increasing? The map $f: t \mapsto |a+tv|_1$ is convex. Let $0<t_1<t_2$. By the chordal slope lemma, we get $$\frac{|a+t_1v|_1 - |a|}{t_1} = \frac{f(t_1) - f(0)}{t_1} \le \frac{f(t_2) - f(0)}{t_2} = \frac{|a+t_2v|_1 - |a|}{t_2}.$$ This implies $\varphi(t_1) \le \varphi(t_2)$. – Analyst Jun 04 '22 at 22:30
  • 1
    Oh, I now see that. I guess I confused myself by not factoring the denominator $t$ into my though process. Now your solution seems having no issues (modulo the typo $t_n$). – Sangchul Lee Jun 04 '22 at 22:38

2 Answers2

2

If $a>0$ then $\|a\|_1 = a_1+\cdots+a_m$ and so the map $f(a)= \|a\|_1$ is differentiable with $Df(a)v = v_1+\cdots + v_m$. Hence $df(a;v) = Df(a)v = v_1+\cdots + v_m$ (the one sided derivative).

In general, if $g(x)=|x|$ we have $df(x,h) = \begin{cases} |h|, & x = 0 \\ (\operatorname{sgn} x)h,& \text{otherwise} \end{cases}$.

copper.hat
  • 178,207
1

Let me brush up your approach to give a valid proof. For each $m = 1, 2, \ldots,$ consider the tail sum

$$ D_m(t) = \sum_{i>m} \frac{|a_i + v_i t| - |a_i|}{t}. $$

By the direct/reverse triangle inequality, we easily find that

$$ \left| D_m(t) \right| \leq \sum_{i>m} \left|\frac{|a_i + v_i t| - |a_i|}{t}\right| \leq \sum_{i>m} \left| v_i \right|. $$

Then it follows that

\begin{align*} \left| D(t) - \sum_{i=1}^{\infty} v_i \right| &\leq \sum_{i=1}^{m} \left| \frac{|a_i + v_i t| - |a_i|}{t} - v_i \right| + \left| D_m(t) \right| + \left| \sum_{i>m} v_i \right| \\ &\leq \sum_{i=1}^{m} \left| \frac{|a_i + v_i t| - |a_i|}{t} - v_i \right| + 2\sum_{i>m} \left| v_i \right| \end{align*}

So, by taking limsup as $t \to 0^+$ and noting that $\frac{|a_i + v_i t| - |a_i|}{t} \to v_i$ as $t \to 0^+$, we get

\begin{align*} \limsup_{t \to 0^+} \left| D(t) - \sum_{i=1}^{\infty} v_i \right| &\leq 2 \sum_{i>m} \left| v_i \right|. \end{align*}

However, since the left-hand side does not depend on the value of $m$, letting $m \to \infty$ shows that the indicated limsup is $0$, proving the desired claim.

Addendum. A quicker solution is available if the dominated convergence theorem (DCT) is applicable: If $a_i \neq 0$ for all $i$ and $v \in \ell^1(\mathbb{N}_1)$, then

\begin{align*} D(t) &= \int_{0}^{1} \sum_{i=1}^{\infty} v_i \operatorname{sgn}(a_i + v_i t s) \, \mathrm{d}s \\ &\quad \xrightarrow[\text{as } t\to 0^+]{\text{DCT}} \ \int_{0}^{1} \sum_{i=1}^{\infty} v_i \operatorname{sgn}(a_i) \, \mathrm{d}s = \sum_{i=1}^{\infty} v_i \operatorname{sgn}(a_i). \end{align*}

Sangchul Lee
  • 181,930