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Let $X_1,\cdots,X_n$ be i.i.d. random variables with density $f$ and let $\hat{f}$ be an estimator of $f$. Is the following inequality direct from the standard properties of expectation and sup-norm?

$$\mathbb{E}\left[ \lVert \hat{f}- \mathbb{E}(\hat{f})\rVert_\infty + \lVert \mathbb{E}(\hat{f})-f\rVert_\infty \right]^2 \leq 2\mathbb{E}\lVert \hat{f}- \mathbb{E}(\hat{f})\rVert_\infty^2 + 2\lVert \mathbb{E}(\hat{f})-f\rVert_\infty^2.$$

This inequality is found in page 44 of Tsybakov's book.

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Since$$(a+b)^2=a^2+2ab+b^2\le a^2+\overbrace{(a^2+b^2)}^{\ge 2ab}+b^2=2a^2+2b^2$$ $$\mathbb{E}\left[ \lVert \hat{f}- \mathbb{E}(\hat{f})\rVert_\infty + \lVert \mathbb{E}(\hat{f})-f\rVert_\infty \right]^2 \leq 2\Bbb E[\lVert \hat{f}- \mathbb{E}(\hat{f})\rVert_\infty]^2 + 2\lVert \mathbb{E}(\hat{f})-f\rVert_\infty^2.$$

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  • $\begingroup$ Can you show me why 2ab need to be less than or equal the sum of squares? This is exactly what I was struggled with. $\endgroup$ – Celine Harumi Mar 24 '19 at 12:48
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    $\begingroup$ Hint:$$a^2+b^2\ge 2ab$$is equivalent to$$a^2+b^2-2ab\ge 0$$and then... $\endgroup$ – Xi'an Mar 24 '19 at 13:53

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