Suppose that $x = [x_1,\dots,x_n]$ is a vector with norm less than or equal to one $\|x\|_2^2 \leq 1$. Let $\alpha \in [0,1]$ and define the following vector
$$y = [x_2+\alpha x_1, \dots, x_n + \alpha x_1]$$
How can I find a non-trivial subset of $x$ that, regardless of the value of $\alpha$, would result in $\|y\|_2^2 \leq C$ where $C$ is a constant that does not depend on $n$?
The above is satisfied when $x_1 = \dots = x_n$. For example, if $x = [1/\sqrt{n}, \dots, 1/\sqrt{n}]$ then, we have $\|y\|^2_2 = \frac{(1+\alpha)^2}{n} \times n = (1+\alpha)^2 \leq 4$. But I'm looking for a larger subset (or all $x$ that satisfy the above conditions) than $x_1 = \dots = x_n$.
You just need the condition $|x_1|\le 1/\sqrt n$ and you get $\|y\|^2\le 4$. In fact, $$ \|y\|^2 = [(n-1)\alpha - 1 -2\alpha] x_1^2 + \sum x_1^2 + 2\alpha x_1\sum x_i $$ but $$ \sum x_i \le n\sqrt{\sum x_i^2/n}\le \sqrt n $$ so $$ \|y\|^2 \le [(n-3)\alpha - 1 ] x_1^2 + 1 + 2\alpha x_1\sqrt n \le 4 $$