The result I am trying to prove that the square, diamond, and Euclidean metric on $\mathbb{R}^n$ induce the same topology. Given a lemma I have access to regarding continuity for dominating metrics, it suffices to verify that each of these metrics dominates the other. My lecture notes feature the following inequalities. For $x = (x_1, \ldots, x_n) \in \mathbb{R}^n$, we have:
\begin{align*}
\max\limits_{i} |x_i| \leq \sum\limits_{i} |x_i| \leq n \cdot \max_{i} |x_i|
\end{align*}
and
\begin{align*}
\max\limits_{i} |x_i| \leq \left(\sum\limits_{i} (x_i)^2 \right)^{1/2} \leq n^{1/2} \max\limits_{i} |x_i|.
\end{align*}
I'm fine with the first inequality, but I'm not fully comfortable with the second. I think the first follows from rewriting $|x_i|$ as $\sqrt{(x_i)^2}$, recognizing the maximal absolute value is only one index, and then summing over all $n$. If I denote that maximal $|x_i|$ by the index $j$, we get:
$$
\max\limits_{i} |x_i| = |x_j| \leq \sqrt{(x_j)^2} = \sqrt{\sum\limits_{i=1}^n (x_i)^2}.
$$
For the second, I think we replace each term of the summand with the maximal square. As the squaring function is nonnegative, the maximal $|x_i|$ will be the maximal $(x_i)^2$ and vice-versa. So we have:
$$
\left(\sum\limits_{i} (x_i)^2 \right)^{1/2} \leq \left(\sum\limits_{i} \max\limits_{i} \{(x_i)^2\} \right)^{1/2} \leq \left(n \max\limits_{i} \{(x_i)^2\} \right)^{1/2} = n^{1/2} \max\limits_{i} |x_i|.
$$
In the first inequality, I replaced $(x_i)^2$ with the maximum of the $(x_i)^2$, letting $i$ vary over $1, \ldots, n$. I summed over $i$ this now constant value and used the fact that the square root function is monotonically increasing. The final line used the fact that the square root of a product is the product of the square root and the square root of the maximum is the maximum of the square root.
How does this look?
Best Answer
To see $$\max\limits_{i} |x_i| \leq \left(\sum\limits_{i} (x_i)^2 \right)^{1/2} \leq n^{1/2} \max\limits_{i} |x_i|$$
I would simply reason as follows:
Indeed for any $i$, $|x_i|=\sqrt{x_i^2}$ and so for the index $j$ where this value is maximal, it is certainly smaller than $\sqrt{\sum_i x_i^2}$ because we add only positive terms in the sum and square root is monotonous. This gives the left $\leq$.
For the other one, just note that we have $x_i^2 \leq \left(\max\limits_i |x_i|\right)^2$ and taking this inequality $n$ times (for each term $i=1,\ldots,n$) we get $\sum\limits_i x_i^2 \leq n \cdot \left(\max\limits_i |x_i|\right)^2$ and then we take the square root on both sides to get the final $\leq$.