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I am currently studying the sequence $(B_n(f))_{n=1}^{\infty}$ of Bernstein Polynomials $$(B_n(f))(x) := \sum_{k=0}^{n}f\left(\frac{k}{n}\right) {n \choose k}x^k(1-x)^{n-k},\quad \text{ where }\quad 0\leq x \leq 1,$$ and came across the following inequality in my analysis textbook: $$\|B_n(f)\|_{\infty} \leq \|f\|_{\infty} \tag{$*$}$$ I already showed that $|B_n(f)| \leq B_n(|f|)$ and $B_n(f) \geq 0$ whenever $f \geq 0$, however, how is the inequality $(*)$ correct? I feel like having shown the other two inequalities I should be able to deduce why $(*)$ is true $($as the author has not provided a proof$)$ but something is not clicking. Any suggestions are welcome.

Note: Showing $|B_n(f)| \leq B_n(|f|)$ and $B_n(f) \geq 0$ whenever $f \geq 0$ was an exercise in the textbook $-$ Showing $(*)$ is not as it was mentioned briefly in the book without proof.

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Note that $|B_n(f)(x)|\leq \|f\|_{\infty} \sum_{k=0}^n {n \choose k} x^k (1-x)^{n-k}$.

Now, $$ 1=1^n=(x+(1-x))^n=\sum_{k=0}^n {n\choose k} x^k(1-x)^{n-k}, $$ by the binomial identity.

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  • $\begingroup$ Does that mean $sup |B_n(f)(x)|\leq \|f\|_{\infty} \sum_{k=0}^n {n \choose k} x^k (1-x)^{n-k}$? $\endgroup$ – Taylor Rendon Nov 17 '20 at 23:46
  • $\begingroup$ I'm just confused as to why this helps prove $\|B_n(f)\|_{\infty} \leq \|f\|_{\infty} \tag{$*$}.$ If you could elaborate more to help me see why this holds using what you said, that would be much appreciated (-:. $\endgroup$ – Taylor Rendon Nov 18 '20 at 0:21
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    $\begingroup$ If you add all the lines together, you now have $|B_n(f)(x)|\leq \|f\|_{\infty}$ for every $x$. $\endgroup$ – WoolierThanThou Nov 18 '20 at 9:00
  • $\begingroup$ Okay, then does this imply that: $|B_{n}(f)(x)| \leq sup |B_{n}(f)(x)| \leq ||f||_{\infty}$? $\endgroup$ – Taylor Rendon Nov 18 '20 at 16:23
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    $\begingroup$ @TaylorRendon No. $\endgroup$ – WoolierThanThou Nov 18 '20 at 21:34

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