Proof that $(t_1, \dots, t_r) \mapsto \sum^{r}_{i=1} | t_i - \alpha_i|^p$ is continuous - Problem with Inequality

Question

Bounty Edit: I already edited the question after some important comments. The questions I have are highlighted below the supposed proof. Any feedback or answer is most welcome.

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Thus, I just found a proof in a book that relies upon the continuity of the following function $\phi : \mathbb{R}^r \to \mathbb{R}$, such that

$$(t_1, \dots, t_r) \mapsto \sum^{r}_{i=1} |t_i - \alpha_i|^p,$$ for $\alpha_i$ arbitrary, and $1 \leq p < \infty$.

Hence, I used this as an opportunity to train my skills on proving continuity and using inequalities (I am quite bad at both things).

Here there is the proof: the inequality I found the most problematic (in the sense that I am kind of unsure how I got it), is the blue one.

Proof: Fix an arbitrary $p$, with $1 \leq p < \infty$. For notational convenience let $\mathbf{t} := (t_1, \dots, t_r)$, and assume that $\mathbf{t}^n \to \mathbf{t}$. That is, for every $\varepsilon >0$, there exists a $M\in \mathbb{N}$ such that for all $n \geq M$, $d_p (\mathbf{t}^n,\mathbf{t})< \varepsilon$. Let $\varepsilon >0$ be arbitrary, and endow $\mathbb{R}^r$. Hence, $$d_p (\mathbf{t}^n,\mathbf{t})< \varepsilon \Longrightarrow | \mathbf{t}^n - \mathbf{t}| < \varepsilon \Longrightarrow \bigg( \sum^r | t_{i}^n - t_i|^p \bigg)^{\frac{1}{p}} < \varepsilon \Longrightarrow \sum^r | t_{i}^n - t_i|^p < \varepsilon^p .$$ Thus, we have the following chain of inequalities that proves the continuity of $phi$: \begin{align} d_p (\phi(\mathbf{t}^n),\phi(\mathbf{t})) & = \bigg( \sum^r | t_{i}^n - \alpha_i|^p - \sum^r | t_{i} - \alpha_i|^p \bigg)^{\frac{1}{p}} \\ & = \bigg( \sum^r | t_{i}^n - \alpha_i|^p - | t_{i} - \alpha_i|^p \bigg)^{\frac{1}{p}}\\ & \color{blue}{\leq \sum^r ( t_{i}^n - \alpha_i) - ( t_{i} - \alpha_i) \hspace{1cm}(*)} \\ & = \sum^r ( t_{i}^n - t_{i})\\ & = \sum^r | t_{i}^n - t_{i}|^p < \varepsilon^p. \end{align}

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Questions:

1. Is it correct?
   I did write the proof again (as you can see in the edit of the question) because Surb's comment below concerning an inequality in the previous attempted proof made me think. Actually, I am not sure I really fixed the issue with this one. Indeed, the blue inequality $(*)$ in the text is the one that makes me think there is something wrong there (i.e. how I get there).
2. If the proof is not correct, how can we prove this result along the same lines I used here?
   How does such a proof runs when correct?
   True, this is not the most efficient way, as pointed out by Ilya's comment below, but I really would like to see how to get the result by means of chains of inequalities.

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Thank you for your time.

Answer 1

There are very many mistakes here:

1. $d_p$ is a distance on $\Bbb R ^r$, but you use it on $\Bbb R$ when you write $d_p (\phi ({\bf t}^n) , \phi ({\bf t}))$
2. the lines above the blue inequality are certainly false: the quantity between brackets may be negative, and if $p=2$ you get imaginary results; you may fix this with a modulus
3. to get to the blue line, you essentially use that $\sum (x_i - y_i)^p = \sum (x_i ^p - y_i ^p)$ which is obviously wrong
4. below the blue inequality you obtain $\sum (t^n_i - t_i)$ which may be negative, in which case you obtain a negative distance (in the left-hand side)
5. on the last line you put a modulus and a power $p$ out of the blue, which is blatantly wrong

I won't look for further errors, any single one mentioned is enough to invalidate your reasoning.

Concerning your second answer, the problem is much easier than expected and it boils down to decomposing your function into a composition of elementary continuous ones:

• the functions ${\bf t} \mapsto t_i$ are all continuous
• addition is continuous, so all ${\bf t} \mapsto t_i - \alpha _i$ are continuous
• the modulus is continuous, so all ${\bf t} \mapsto |t_i - \alpha _i|$ are continuous
• powers are continuous, so all ${\bf t} \mapsto |t_i - \alpha _i|^p$ are continuous
• again, addition is continuous, so ${\bf t} \mapsto \sum \limits _{i=1} ^r |t_i - \alpha _i|^p$ is continuous
• again, powers are continuous so ${\bf t} \mapsto \left( \sum \limits _{i=1} ^r |t_i - \alpha _i|^p \right) ^{\frac 1 p}$ is continuous.

Therefore, your function is continuous as a composition of elementary continuous functions.