# A technical lemma on linear combinations.

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?

• 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 $r_i \rightarrow 0$. Mar 4, 2023 at 18:07
• Does this answer your question? Inequality for norm of linear combination of linearly independent vectors Mar 4, 2023 at 18:48
• 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 $\operatorname{span}\{x_1,\ldots, x_n\}$? Mar 4, 2023 at 19:00
• @mechanotroid You should tranform your comment into an answer that I would be happy to upvote. Mar 5, 2023 at 5:18
• Another track : math.stackexchange.com/q/165041/305862 Mar 5, 2023 at 7:43

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