Lemma Let $\{x_1,\dots, x_n\}$ be a linearly independent set of vectors in a normed space $X$. Then there is a number $c>0$ such that for every choice of scalars $\alpha_1,\dots,\alpha_n$ we have $$\lVert \alpha_1x_1+\cdots+\alpha_nx_n\rVert \ge c\left(\lvert\alpha_1\rvert+\dots+\lvert \alpha_n\rvert\right)\quad (c>0).\tag1$$

Comment: We all know that this is the notorious Lemma 2.4-1 of the book Introductory to Functional Analysis by Kreyszig. Before formalizing the question I searched the forum, but no answer fully satisfied my doubts, the answers I found were alternative proofs or not very exhaustive explanations. Please, I would like you to give me as detailed explanations as possible to the questions, evenif the are trivial, that I am going to ask you. I rewrite the proof and I hope that some of you can clarify my doubts also for those who will come after me.

Proof We write $s=\lvert\alpha_1\rvert+\cdots+\lvert\alpha_n \rvert$If $s=0$, all $\alpha_j$ are zero, so $(1)$ holds for any $c$. Let $s>0$. Then $(1)$ is equivalent to the inequality which we obtain from $(1)$ by dividing by $s$ and writing $\beta_j=\alpha_j/s$, that is, $$\tag2 \lVert\beta_1x_1+\cdots+\beta_n x_n\rVert\ge c\qquad \left( \sum_{j=1}^n \lvert\beta_j\rvert=1\right).$$ Hence it suffices to prove the existence of a $c>0$ such that $(2)$ holds for every $n-tuple$ of scalars $\beta_1,\dots, \beta_n$ with $\sum\lvert \beta_j \rvert=1$. Suppose that this is false. Then there exists a sequence $\{y_m\}$ of vectors $$y_m=\beta_1^{(m)}x_1+\cdots+\beta_n^{(m)}x_n\qquad \left(\sum_{j=1}^n\lvert \beta_j^{(m)}\rvert=1\right)$$ such that $$\lVert y_m \rVert\to 0\quad\text{as}\; m\to \infty$$

Question 1 Why this fact deny the hypothesis of the existence of $c$?

Now we reason as follows. Sice $\sum\lvert \beta_j^{(m)}\rvert=1$, we have $\lvert\beta_j^{(m)}\rvert\le 1$. Hence for each fixed $j$ the sequence $$\left(\beta_j^{(m)}\right)=\left(\beta_j^{(1)},\beta_j^{(2)},\dots\right)$$ is bounded. Consequently, by the Bolzano-Weierstrass theorem, $\left(\beta_1^{(m)}\right)$ has a converget subsequence.

Let $\beta_1$ denote the limit of that subsequence, and let $\left(y_{1,m}\right)$ denote the corresponding subsequence, and let $\left(y_{1,m}\right)$ denote the corresponding subsequence of $\left( y_m\right)$. By the same argument, $\left(y_{1,m}\right)$ has a subsequence $\left(y_{2,m}\right)$ for which the corresponding subsequence of scalars $\beta_2^{(m)}$ converges; let $\beta_2$ denote the limit. Continuing in this way, after $n$ steps we obtain a subsequence $(y_{n,m})=(y_{n,1},y_{n,2},\dots)$ of $(y_m)$ whose terms are of the form $$y_{n,m}=\sum_{j=1}^n\gamma^{(m)}x_j\qquad\left(\sum_{j=1}^n\lvert\gamma_j^{(m)}\rvert=1\right)$$ with scalars $\gamma_j^{(m)}$ satisfyng $\gamma_j^{(m)}\to\beta_j$ as $m\to \infty$. Hence, as $m\to\infty,$ $$y_{n,m}\to y=\sum_{j=1}^n\beta_jx_j$$ where $\sum_\lvert\beta_j\rvert=1$, so that not all $\beta_j$ can be zero.

Question 2. I didn't understand the highlighted part above, could someone please explain the details to me?

  • $\begingroup$ Re qn 1, they are doing a proof by contradiction. If there is no $c$ that bounds below, then we can find a sequence of vectors with coefficient sum 1 and norm that tends to 0 (since the norm is not bounded below). $\quad$ Maybe you're confused by the setup? If so, think about investigating if a set of real numbers $r_i$ is bounded away from 0. Either we have $|r| \geq c > 0$, or some subsequence $r_i \rightarrow 0$. $\endgroup$
    – Calvin Lin
    Mar 4, 2023 at 18:07
  • $\begingroup$ Does this answer your question? Inequality for norm of linear combination of linearly independent vectors $\endgroup$
    – Jean Marie
    Mar 4, 2023 at 18:48
  • 3
    $\begingroup$ Doesn't this just follow from the fact that the norms $\|\cdot\|$ and $\sum_{k=1}^n \alpha_kx_k \mapsto \sum_{k=1}^n |\alpha_k|$ are equivalent on the finite-dimensional vector space $\operatorname{span}\{x_1,\ldots, x_n\}$? $\endgroup$ Mar 4, 2023 at 19:00
  • $\begingroup$ @mechanotroid You should tranform your comment into an answer that I would be happy to upvote. $\endgroup$
    – Jean Marie
    Mar 5, 2023 at 5:18
  • $\begingroup$ Another track : math.stackexchange.com/q/165041/305862 $\endgroup$
    – Jean Marie
    Mar 5, 2023 at 7:43

1 Answer 1


$$\lVert y_m \rVert\to 0\quad\text{as}\; m\to \infty$$ precisely means that for any $c>0$, there exist some $m\in \mathbb{N}$ such that $\lVert y_M \rVert<c$ for all $M>m$.

For the 2nd part, the trick here is to first focus only on the first coordinate $\beta_1^{(m)}$. All the terms are bounded by 1, so it has a convergent subsequence, which converges to $\beta_{1}$. Now with your fixed first coordinate, you move on to the 2nd coordinate $\beta_2^{(m)}$. Following the same logic, the terms of 2nd coordinate has a convergent subsequence converging to $\beta_{2}$. Repeat the process till the last coordinate, and finally you will arrive at a sequence $(y_{n,m})=(y_{n,1},y_{n,2},\dots)$ converging to $(\beta_1,\beta_2,\dots)$. Now $ $$y_{n,m}\to y=\sum_{j=1}^n\beta_jx_j$$ where $$\sum_\lvert\beta_j\rvert=1$. As atleast one $\beta_j$ is non zero so $\lVert y \rVert>0$. (Note that all the $x_j$'s are nonzero as they are linearly independent).

Hence it contradicts our assumption that $$\lVert y_m \rVert\to 0\quad\text{as}\; m\to \infty$$ Hence the proof.


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