Passing from induction to $\infty$

Question

Somehow, the operation of passing to the limit after I have shown that something is true by induction for each natural number $n$ troubles me each time. I know there are instances where one cannot deduce a statement is true for $\infty$ if it holds for each $n \in \mathbb{N}$, and sometimes one can.

Here is one instance where I am not sure whether my argument is solid:

Suppose $X$ is a Banach space with norm $\| \cdot \|$. I would like to show that the triangle inequality for finite linear combinations extends to series. So, by induction and by properties of the norm we have

$$\begin{equation} \bigg\|\sum^n_{k = 1} a_k \bigg\| \;\;\leq \;\;\sum_{k = 1}^n \|a_k\| \end{equation}$$

holds for each natural number $n$. I would now argue that I can immediately pass to the limit and deduce directly that

$$\begin{equation} \bigg\|\sum^\infty_{k = 1} a_k\bigg\| \;\;\leq \;\;\sum_{k = 1}^\infty \|a_k\| \end{equation} $$

because of a property of sequences of real numbers, which says that if $(c_n)_{n \in \mathbb{N}}$ and $(d_n)_{n \in \mathbb{N}}$ are sequences of real numbers such that

\begin{equation} 0 \leq c_n \ \leq d_n \quad \text{ for all } n \in \mathbb{N}. \end{equation}

Then it follows that

\begin{equation} \lim_{n \to \infty} c_n \leq \lim_{n \to \infty} d_n \end{equation}

(I can use this by taking $c_n = \| \sum_{k = 1}^n a_k \|$ and $d_n = \sum_{k = 1}^n \|a_k\|$)

Is this reasoning correct ? I am not sure ..

For example, one of the issues I have with my argument is the following:

When I replace $n$ by $\infty$ in the expression $\|\sum_{k = 1}^n a_k \|$ then I might make a statement that is ill-defined, because $\|\sum_{k = 1}^\infty a_k \|$ might not exist, whereas the expression $\sum_{k = 1}^\infty \|a_k\|$ always has a value in $[0,\infty]$ (since $s_n = \sum_{k = 1}^n \|a_k\|$ is a sequence that is monotone).

Is this an issue? Or is the statement $\|\sum_{k = 1}^\infty a_k \| \leq \sum_{k = 1}^\infty \|a_k\|$ simply vacuously true in this case?

If it is an issue, how can I rectify the argument?

Answer 1

What your argument shows is that, for any sequence $a_k$ in $X$, we always have $$\bigg\|\sum_{k=1}^na_k\bigg\|\leq\sum_{k=1}^n\|a_k\|$$ for any $n$, and therefore $$\lim_{n\to\infty}\bigg\|\sum_{k=1}^na_k\bigg\|\leq\lim_{n\to\infty}\sum_{k=1}^n\|a_k\|\;\overset{\text{ by definition }}{=}\;\sum_{k=1}^\infty\|a_k\|$$

But we still need to show that, for any convergent series $\sum_{k=1}^\infty a_k$ in $X$, $$\bigg\|\sum_{k=1}^\infty a_k\bigg\|=\lim_{n\to\infty}\bigg\|\sum_{k=1}^na_k\bigg\|$$ in order to conclude that $$\bigg\|\sum_{k=1}^\infty a_k\bigg\|\leq \sum_{k=1}^\infty\|a_k\|.$$ (Of course, if $\sum_{k=1}^\infty a_k$ doesn't converge, then it makes no sense to talk about its norm in the first place, so we may as well assume it converges.)

The fact that $$\bigg\|\sum_{k=1}^\infty a_k\bigg\|=\lim_{n\to\infty}\bigg\|\sum_{k=1}^na_k\bigg\|$$ follows from the continuity of the norm function $\|\cdot\|:X\to\mathbb{R}$. Letting $$S_n=\sum_{k=1}^n a_k,$$ we have that $$\sum_{k=1}^\infty a_k=\lim_{n\to\infty} S_n.$$ For any topological spaces $A$ and $B$, continuous function $f:A\to B$, and convergent sequence $x_n\to x$ in $A$, we have that $f(x_n)\to f(x)$ in $B$. Thus, because the norm function $\|\cdot\|$ is continuous, we will get that $$\|\lim_{n\to\infty} S_n\|=\lim_{n\to\infty}\|S_n||$$ or in other words $$\bigg\|\sum_{k=1}^\infty a_k\bigg\|=\lim_{n\to\infty}\bigg\|\sum_{k=1}^na_k\bigg\|.$$

The reason the norm function is continuous is simply that the norm itself is what defines the topology on $X$. That is, the definition of $x_n\to x$ in $X$ is that $\|x_n-x\|\to 0$ in $\mathbb{R}$, and because the reverse triangle inequality tells us that $$\big|\|x\|-\|x_n\|\big|\leq\|x_n-x\|$$ we have that $$\lim_{n\to\infty}\big|\|x\|-\|x_n\|\big|=0$$ and hence $\lim_{n\to\infty}\|x_n\|=\|x\|$.

Answer 2

First, let me mention that it's a good thing that you have misgivings, because something which is true for every finite step need not be true "at $\infty$"; that is, induction tells you something about "every finite stage", but not necessarily something in the "infinite" case. The standard example is we can prove that for every $n$, the set of natural numbers less than or equal to $n$ is finite; but this is no longer true if we "let $n$ go to infinity."

Moreover, you can't really go from $c_n\leq d_n$ to $\lim\limits_{n\to\infty}c_n\leq\lim\limits_{n\to\infty}d_n$. For one thing, one (or both) of the limits may fail to exist.

If both limits exist, you can deduce the result. Here's one way to do it:

Suppose $c_n\leq d_n$ for all $n$, and that each of $c=\lim\limits_{n\to\infty}c_n$ and $d=\lim\limits_{n\to\infty}d_n$ exists in the extended reals (we allow $\infty$ and $-\infty$ as limits). I claim that $c\leq d$.

Indeed, assume that $d\lt c$. Then there exists $\epsilon\gt 0$ such that $d+2\epsilon\lt c$. Since $\lim\limits_{n\to\infty} d_n = d$, there exists $N\gt 0$ such that for all $n\geq N$, $|d_n-d|\lt\epsilon$. In particular, $d_n\in(d-\epsilon,d+\epsilon)$. Therefore, for all $n\geq N$ we have $c_n\leq d_n\lt d+\epsilon\lt c-\epsilon$, so $|c-c_n|\gt \epsilon$ for all $n\geq N$. This contradicts the fact that $\lim\limits_{n\to\infty}c_n=c$. The contradiction arises from the assumption that $d\lt c$, so we conclude that $c\leq d$.

From this, since $d_n = \sum_{i=1}^n\lVert a_i\rVert$ is an increasing sequence, it converges (in the extended reals). So provided that $c_n=\lVert\sum_{i=1}^na_i\rVert$ converges, you can conclude your inequality.

In fact, you can extend this to the following:

Theorem. Suppose that $\{c_n\}$ and $\{d_n\}$ are sequences of real numbers such that $c_n\leq d_n$ for every $n$. Then $$\limsup c_n \leq \limsup d_n.$$

Proof. Assume $\limsup d_n = d \lt c =\limsup c_n$. Since $\limsup c_n$ is the supremum of all limit points of subsequences of $c_n$, there is a limit point $\ell$, $d\lt \ell\leq c_n$ and a subsequence $c_{n_k}$ of $\{c_n\}$ such that $\lim\limits_{k\to\infty}c_{n_k}=\ell$. In particular, there is an $\epsilon\gt 0$ such that for all sufficiently large $k$ we have $$d+\epsilon \lt c_{n_k}\leq d_{n_k}.$$ On the other hand, $\{d_{n_k}\}$ is either unbounded (in which case it has a subsequence converging to $\infty$) or it contains a convergent subsequence. In either case, $\{d_{n_k}\}$ has a limit point that is greater than $d+\epsilon$, so $\limsup d_{n_k}\geq d+\epsilon\gt d$. But this gives $d\lt \limsup d_{n_k}\leq \limsup d_n = d$, a contradiction.

Similarly, we can conclude that $\liminf c_n \leq \liminf d_n$ under the same hypotheses.

Answer 3

Note: I am assuming these sums in fact exist. If they fail to, the statement is meaningless.

You are not deducing anything about the $\infty$ case, which isn't even well-defined. The statement $$\left\|\sum\limits_{k=1}^\infty a_k\right\|\leq \sum\limits_{k=1}^\infty \|a_k\|$$ is defined as follows:

Let $x$ be the unique real number such that for all $\epsilon>0$ there exists some $N\in\mathbb N$ such that $$n\geq N\implies \left\|x-\sum\limits_{k=1}^n a_k\right\|<\epsilon$$ and let $y$ be the unique real number such that for all $\epsilon>0$ there exists some $N\in\mathbb N$ such that $$n\geq N\implies \left\|y-\sum\limits_{k=1}^n \|a_k\|\right\|<\epsilon.$$ Then $x\leq y$.

This makes no reference to $\infty$. Indeed, it works perfectly well to show that $$\left\|\sum\limits_{k=1}^n a_k\right\|\leq \sum\limits_{k=1}^n \|a_k\|$$ for all $n$ since this means that for any $\epsilon>0$ we have some $N$ such that if $n\geq N$ then $$x\leq y+\|x-y\|\leq y+\left\|\sum\limits_{k=1}^n a_k\right\|-\sum\limits_{k=1}^n \|a_k\|-\left\|x-\sum\limits_{k=1}^n a_k\right\|-\left\|y-\sum\limits_{k=1}^n \|a_k\|\right\|<y+2\epsilon$$ so we must have $x\leq y$.

Answer 4

The problem disappears as soon as you correctly formulate what you want to prove. The correct formulation is: If $\sum_{k=1}^\infty a_k$ converges then $\|\sum_{k=1}^\infty a_k\|\le \sum_{k=1}^\infty \|a_k\|$. The precise argument to prove this result is true is obtained by following the definition of series as a limit of partial sums and using the finite triangle inequality, and then pass to the limit (basically as you indicate above). But now you don't need to worry about the inequality 'holding vacuously' (a dangerous term to use).

Also, a general comment about how proofs are done, or rather how they are not done. Passing to the limit does not mean replacing $n$ by $\infty$. It means using theorems about (in the case above inequalities between elements of sequences and their) limits. Replacing $n$ by $\infty $ is meaningless. This might be the reason why you feel uncomfortable doing it.

Answer 5

And supposing that $\sum\limits_{k=1}^\infty\|a_k\|$ converges, $\|\sum\limits_{k=1}^\infty a_k\|$ makes sense, since X is a Banach space (and every absolute convergent serie is convergent).