# Rudin Theorem $3.4 (b)$

Rudin left this proof as an exercise, so I would appreciate if someone could look over my attempt.

(b) Suppose $$\{x_n\}$$, $$\{y_n\}$$ are sequences in $$\mathbb{R}^k$$, $$\{\beta_n\}$$ is a sequence of real numbers, and $$x_n \to x$$, $$y_n \to y$$, $$\beta_n \to \beta$$. Then \begin{align*} \lim\limits_{n \to \infty} (x_n + y_n) & = x + y \\ \lim\limits_{n \to \infty} x_n \cdot y_n & = x \cdot y \\ \lim\limits_{n \to \infty} \beta_n x_n & = \beta x. \end{align*}

Here is my attempt. I'm going to stick to the notation used by Rudin.

Let $$x_n = (a_{1,n}, \ldots, a_{k,n})$$, $$y_n = (b_{1,n} \ldots, b_{k,n})$$, $$\beta_n = (\beta_1, \ldots, \beta_k)$$, $$x = (a_1, \ldots, a_n)$$, $$y = (b_1, \ldots, b_n)$$. Then for any $$n$$, we have: $$x_n + y_n = (a_{1,n} + b_{1,n}, \ldots, a_{k,n} + b_{k,n})$$ Since $$x_n \to x$$, by part (a), the sequence converges component-wise, so $$a_{j,n} \to a_j$$ for each $$j$$. Since $$y_n \to y$$, we have $$b_{j,n} \to b_j$$ for each $$j$$. So for each $$j$$, $$a_{j,n} + b_{j,n} \to a_j + b_j$$ by Theorem 3.3(a) (sum of convergent sequences). So, by part (a) again, since $$x_n + y_n$$ converges componentwise, we have $$x_n + y_n \to (a_1 + b_1, \ldots, a_k + b_k) = (a_1, \ldots, a_k) + (b_1, \ldots, b_k) = x + y.$$ Moving to the second limit. For any $$n$$, we have $$x_n \cdot y_n = (a_{1,n}, \ldots, a_{k,n}) \cdot (b_{1,n}, \ldots, a_{k,n}) = \sum\limits_{i=1}^k a_{i,n} b_{i,n}.$$ Again, as $$x_n \to x$$, we have $$a_{j,n} \to a_j$$, and as $$y_n \to y$$, we have $$b_{j,n} \to b_j$$. By Theorem 3.3, products of convergent sequences convergence, so for each $$j$$, we have $$a_{j,n} b_{j,n}$$ converges to $$a_j b_j$$. By Theorem 3.3 again and induction on $$k$$, we have \begin{align*} \sum\limits_{i=1}^k a_{i,n} b_{i,n} \to \sum\limits_{i=1}^k a_i b_i = (a_1, \ldots, a_k) \cdot (b_1, \ldots, b_k) = x \cdot y. \end{align*} Moving now to the third limit. For any $$n$$, we have $$\beta_n x_n = (\beta_n a_{1,n}, \ldots, \beta_{n} a_{k,n}).$$ By Theorem 3.3, for each $$j$$, we have $$\beta_n a_{j,n} \to \beta a_j$$ (product of convergent sequences). So $$\beta_n x_n$$ converges component-wise, so it converges, and we have $$\beta_n x_n \to (\beta a_1, \ldots, \beta a_k) = \beta (a_1, \ldots, a_k) = \beta x.$$

How do these look?

• You wrote $x = (a_1, \ldots, a_n)$. Should it be $x = (a_1, \ldots, a_k)$? Commented Jun 5, 2021 at 1:21
• The $\beta_n$ were supposed to be real numbers (scalars).
– plop
Commented Jun 5, 2021 at 1:26
• One thing that is not wrong, but can save you some writing and improve clarity is to avoid expressing things in coordinates. If you want to prove $\lim_{n\to\infty}(x_n+y_n)=x+y$, you only need to subtract and take the norm $\|(x_n+y_n)-(x+y)\|=\|(x_n-x)+(y_n-y)\|$. Then apply the triangle inequality $\|(x_n-x)+(y_n-y)\|\leq \|x_n-x\|+\|y_n-y\|$. Since $\lim_{n}x_n=x$ and $\lim_ny_n=y$, there is $N$ such that for all $n>N$ we have $\|x_n-x\|<\epsilon/2$ and $\|y_n-y\|<\epsilon/2$. Therefore, $\|(x_n+y_n)-(x+y)\|<\epsilon$
– plop
Commented Jun 5, 2021 at 1:51
• @JeremyS the coordinates of $x_n$ should all be multiplied by the same $\beta_n$. You are still multiplying them individually by the numbers $\beta_1, \dots, \beta_k$ instead, which is a mistake. Commented Jun 5, 2021 at 2:37
• It still shows $\beta_n = (\beta_1, \dots, \beta_k)$, even though this is not the case, at the beginning of the proof, but otherwise it looks fine. Unrelated, but as a tip, if you want a response from someone that isn’t the OP in the comments, tag them with an @ or else they will not be notified. I just happened to revisit this post and noticed you replied. Commented Jun 5, 2021 at 2:57

These all look fine. The proofs are long because you have relied on the facts of convergence from $$\mathbb R^1$$, as well as finite-dimensionality, but the proofs are correct. These proofs wouldn't work, however, in infinite-dimensional normed vector spaces, which you will undoubtedly soon study if you continue in analysis.

The relevant idea here that simplifies all the proofs, and generalizes to infinite dimensions, is that the vector space structure is compatible with the distance. The Euclidean spaces are all inner product spaces with induced norm $$|x|=\sqrt{x\cdot x}$$, so the only necessary data is the vector space structure and the norm (or inner product as necessary). For example,

1. Proof of the first limit:

$$|(x_n+y_n)-(x+y)| \le |x_n-x| + |y_n-y| \to 0.$$

1. Proof of the second limit: \begin{align*} |x_n\cdot y_n - x\cdot y| &= |(x_n-x)\cdot y_n + x\cdot (y_n -y)|\\ &\le |(x_n-x)\cdot y_n| + |x\cdot(y_n-y)|\\ &\text{Now apply Cauchy-Schwarz to finish}\\ &\le |x_n-x|\,|y_n| + |x|\,|y_n-y| \to 0. \end{align*} (Small comment, this idea of adding and subtracting a term like this crops up all the time when dealing with "products", including the dot product, and more generally inner products. It doesn't have a name as far as I know, but let's call it the bilinear trick.)

2. Proof of the last limit (a product is involved, so think about the bilinear trick): \begin{align*} |\beta_nx_n - \beta x| &= |(\beta_n-\beta)x_n + \beta(x_n-x)| \\ &\le |\beta_n-\beta|\,|x_n| + |\beta|\,|x_n-x| \to 0. \end{align*}