# How is the inequality $\|B_n(f)\|_{\infty} \leq \|f\|_{\infty}$ correct?

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.

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.
• Does that mean $sup |B_n(f)(x)|\leq \|f\|_{\infty} \sum_{k=0}^n {n \choose k} x^k (1-x)^{n-k}$? – Taylor Rendon Nov 17 '20 at 23:46
• 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 (-:. – Taylor Rendon Nov 18 '20 at 0:21
• If you add all the lines together, you now have $|B_n(f)(x)|\leq \|f\|_{\infty}$ for every $x$. – WoolierThanThou Nov 18 '20 at 9:00
• Okay, then does this imply that: $|B_{n}(f)(x)| \leq sup |B_{n}(f)(x)| \leq ||f||_{\infty}$? – Taylor Rendon Nov 18 '20 at 16:23