# Questions about proof of $\lim x_n = a, \lim y_n = b\implies \lim x_n+y_n = a+b$ in a normed vector space

I need to prove that, in a normed vector space $E$, we have:

$$\lim x_n = a, \lim y_n = b\implies \lim (x_n+y_n) = a+b$$

and:

$$\lim\lambda_n = \lambda, \lim x_n = a \implies \lim \lambda_n\cdot x_n = \lambda\cdot a$$

where $\lambda_n$ is a sequence in $\mathbb{R}$.

The proof of this fact is very different than I thought It would be:

Let $f,g:\overline{P}\to E$ defined by $f(\frac{1}{n})=x_n, f(0) = a, g(\frac{1}{n}) = y_n, g(0) = b$. Then $f+g(\frac{1}{n}) = x_n + y_n$ and $f+g(0) = a+b$. The fact that $\lim x_n = a$ and $\lim y_n = b$ assert that $f$ and $g$ are continuos, then, we can use the theorem (proved later in the book) that:

$$\lim x_n = a \iff f:\overline{P}\to M \mbox{ is continuos}$$ Where $\overline{P} = \{0, 1, \frac{1}{2}, \cdots, \frac{1}{n}\}$

I'd prove it like this:

$\lim x_n = a, \lim y_n = b \implies \forall \epsilon>0, \exists n_1, n_2\in\mathbb{N}$ such that:

$$n>n_1\implies |x_n-a|<\frac{\epsilon}{2}$$ $$n>n_2\implies |y_n-b|<\frac{\epsilon}{2}$$

Then choose $n_3 = max\{n_1, n_2\}$ and we have:

$$n>n_3 \implies$$ $$|x_n-a|<\frac{\epsilon}{2}$$ $$|y_n-b|<\frac{\epsilon}{2}$$

Use: $|x_n-a-(y_n-b)|\le|x_n-a|+|y_n+b|$ so

$$n>n_3\implies |x_n-a-(y_n-b)|<\epsilon$$

Which complete my proof for the sum. Is it right? Do I have to use the fact that $E$ is a normed vector space, or I can just assume that $||$ is my metric? I think I'd have to assume $E$ is a normed vector space in the proof for the product. How do I prove it my way, and how do I prove it using the theorem above?

## 1 Answer

Is it right?

Yes, both proofs are correct. They use different approaches: the proposed is topological and yours is anlytical.

Do I have to use the fact that $E$ is a normed vector space, or I can just assume that $||$ is my metric?

No, you cannot, because the inequality $$|x_n-a-(y_n-b)|\le|x_n-a|+|y_n-b|$$ which you used is an inequality for norms, not the triangle inequality as $$d(x_n+y_n, a+b)\le d(x_n+y_n, x_n+b)+ d(x_n+b, a+b)$$ for the metric $d$. You may simplify the RHS of this inequality provided the metric $d$ is (translation) invariant, that is $$d(x+y,x+z)=d(y,z)\mbox{ for each }x,y,z\in E,$$ which makes the function $d(x,0)$ more similar to norm.

I think I'd have to assume $E$ is a normed vector space in the proof for the product.

For the your way proof yes, for the proof using the theorem it suffices for $E$ to be a topological vector space.

How do I prove it my way,

Since the sequences $\{\lambda_n\}$ and $\{x_n\}$ converge, they are bounded, that is there exist numbers $\Lambda, M>0$ such that $|\lambda_n|\le\Lambda$ and $|x_n|\le M$ for all $n$. Next we can proceed similarly

$\lim\lambda_n =\lambda, \lim x_n = x \implies \forall \epsilon>0, \exists n_1, n_2\in\mathbb{N}$ such that:

$$n>n_1\implies |\lambda_n -\lambda |<\frac{\epsilon}{2M}$$ $$n>n_2\implies |x_n - x|<\frac{\epsilon}{2\Lambda}$$

Then choose $n_3 = max\{n_1, n_2\}$ and we have:

$$n>n_3 \implies$$ $$|\lambda_n -\lambda |\frac{\epsilon}{2M}$$ $$|x_n-x|<\frac{\epsilon}{2\Lambda}$$

Use: $|\lambda_n x_n-\lambda x|\le|\lambda_n x_n-\lambda_n x|+|\lambda_n x-\lambda x|= |\lambda_n| |x_n- x|+|\lambda_n-\lambda|| x|<\Lambda\frac{\epsilon}{2\Lambda}+\left(\frac{\epsilon}{2M}\right)M =\epsilon$, so

$$n>n_3\implies |\lambda_n x_n-\lambda x|<\epsilon$$

Which complete your way proof for the product.

*

Again, we had to assume that $E$ is a normed space because this time we used not only the triangle inequality for a metric, but a multiplicativity of a norm. We can adapt this proof for a metric $d$, provided $d$ is homogeneous, that is $$d(\lambda x, \lambda y)=|\lambda|d(x,y)\mbox{ for each }\lambda\in R\mbox{ and }x,y\in E,$$ which makes the function $d(x,0)$ more similar to norm. If $d$ is both invariant and homogeneous, then the function $d(x,0)$ is norm.

and how do I prove it using the theorem above?

Let $f:\overline{P}\to\Bbb R$, $g:\overline{P}\to E$ defined by $f(\frac{1}{n})=\lambda_n, f(0) =\lambda, g(\frac{1}{n}) =x_n, g(0) = x$. Then $f(\frac{1}{n})\cdot g(\frac{1}{n}) =\lambda x_n$ and $f(0)\cdot g(0) =\lambda x$. The fact that $\lim \lambda_n =\lambda$ and $\lim x_n = x$ assert that $f$ and $g$ are continuous. Since $E$ is a topological vector space, the product $f\cdot g$ is continuous too, so the theorem (proved later in the book) implies that $\lim \lambda_n x_n=\lambda x$.