This is exercise 6, part b) in chapter 5 out of Folland's Real Analysis textbook.

Let $X$ be a finite dimensional vector space (over $\mathbb{R}$) with basis $\{e_i\}_{i = 1}^{n}$. Equip $X$ with the following norm: $||\sum_{j=1}^{n}a_je_j||_1 = \sum_{j=1}^{n}|a_j|$

Next, define the map $$p: \mathbb{R}^n \longrightarrow X$$ $$(a_1, ... ,a_n) \mapsto \sum_{j=1}^{n}a_je_j$$

I am interested in showing that the map $p$ is continuous when $\mathbb{R}^n$ is equipped with the usual Euclidean topology and $X$ is equipped with the topology induced by $|| \cdot ||_1$. I would appreciate if anyone could verify my proof.

Firstly, it is clear that the map $p$ is linear. Next, since both the domain and range of $p$ are metric spaces (induced by the norm), it suffices to show sequential continuity.

I will use superscripts to denote the sequence index and subscripts to denote the number component of the $n$-tuple.

Let $\{x^k\}$ be a sequence in $\mathbb{R}^n$ that converges to $x = (x_1, ..., x_n)$ in the Euclidean norm. It is then a standard fact that each component $x_j^k$ converges to $x_j$ in $\mathbb{R}$ for $1 \leq j \leq n$.

Thus, given $\epsilon > 0$, for each $j \in \{1, ..., n\}$, there is a natural number $N_j$ such that $|x_j^k - x_j| < \frac{\epsilon}{n}$ whenever $k \geq N_j$. Define $N:= \max\{N_j\}_{j=1}^n$.

Then for $k \geq N$, we have that $||p(x^k) - p(x) ||_1 = ||p(x^k - x)||_1$ (since $p$ is linear) = $||p((x_1^k - x_1, ... , x_n^k - x_n))||_1 = ||\sum_{j=1}^{n}(x_j^k - x_j)e_j ||_1 = \sum_{j=1}^{n}|x_j^k - x_j| < \sum_{j=1}^{n}\frac{\epsilon}{n} = \epsilon$

Thus, $p(x^k) \longrightarrow p(x)$ in $X$.

Does this check out? Also, since $p$ is linear it would also suffice to show that it is bounded for then continutity would follow. However, I could not find an easy way to show this.


Your proof of continuity is fine!

For showing boundedness, note that we must find a constant $C$ such that $C\|v\| \geq |p(v)|_1$ for all $v \in \mathbb R^n$.

For this,note that if $v = (v_1,...,v_n)$ then $\sum |v_i| = |p(v)|_1$ by the definition. Note that $||v|| = \sqrt{\sum v_i^2}$ is the Euclidean distance.

Now, the famous RMS-AM inequality tells us that for any real quantities $v_i$, we have $$\sqrt{\frac{\sum v_i^2}{n}} \geq \frac{\sum |v_i|}{n}$$

Which gives a constant $C =\sqrt n$.

  • $\begingroup$ That's awesome! I did not know about these inequalities. It makes the proof one line. Thank you for the response and verification. $\endgroup$ – Nicholas Roberts Aug 22 '19 at 4:19
  • $\begingroup$ You are welcome! You will deal with Holder inequality, Jensen inequality etc. in more generality when you go further in the subject, since these inequalities are often important in showing these "norm" comparison inequalities. In fact, the whole of $L^p$ spaces run on the Holder inequalities/ the Cauchy Schwarz inequality. $\endgroup$ – Teresa Lisbon Aug 22 '19 at 4:20
  • $\begingroup$ Yes, I'm familiar with these already having gone through two semesters of graduate analysis. Is the RMS-AM inequality a special case of any of the aforementioned ones? $\endgroup$ – Nicholas Roberts Aug 22 '19 at 4:22
  • 1
    $\begingroup$ Yes, I believe you can derive it from Cauchy Schwarz, though I must verify this. $\endgroup$ – Teresa Lisbon Aug 22 '19 at 4:23

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