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I have from a problem im solving and so far I have that $|f_i(x) - f_i(y)| < \epsilon / \sqrt{n}$ where $\epsilon , n >0$

I'm now trying to use this inequality here:

$$\|f(x) - f(y) \|_2 = [ \sum_i^n (f_i(x) - f_i(y))^2 ] ^{1/2} < \left ( \sum_i^n (\epsilon / \sqrt{n} )^2 \right )^{1/2} = \epsilon$$

This is the result I want, but I am not sure if I can just apply the inequality directly under the square root of the sum.

Furthermore, the book I'm looking at directly wrote:

$$[ \sum_i^n (f_i(x) - f_i(y))^2 ] ^{1/2} \leq n^{1/2} \text{max}_i | f_i(x) - f_i(y) | < n^{1/2} \epsilon / \sqrt{n} = \epsilon$$

I don't understand what they have done here, so I am asking:

  • If someone could please explain what they've done
  • If what I've done is at all correct, if not how I can correct it

many thanks

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1 Answers1

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Th first inequality $$[ \sum_i^n (f_i(x) - f_i(y))^2 ] ^{1/2} < \left ( \sum_i^n (\epsilon / \sqrt{n} )^2 \right )^{1/2}$$ is justified as the square root function is increasing.

The second one $$[ \sum_i^n (f_i(x) - f_i(y))^2 ] ^{1/2} \leq n^{1/2} \text{max}_i | f_i(x) - f_i(y) | $$ is also justified as $$0 \le (f_i(x) - f_i(y))^2 \le (\text{max}_i | f_i(x) - f_i(y) |)^2$$ for $1 \le i \le n$ and again because the square root is an increasing function.

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