Given a function $f(a) = a^4 +a^3 +2a^2 +a+1$ . How can I show that it is a coercive function? I am having a hard time understanding how to calculate the norm of $f$ in order to show that f is coercive. Can someone help me? Thanks!
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What is the definition of coercive that you're using? Is it as simple as $f(x) \to \infty$ as $|x| \to \infty$? – Theo Bendit Apr 22 '21 at 04:58
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Yes, that is the same definition I'm using. – Blake Apr 22 '21 at 05:00
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So, I take it it's not sufficient to observe that $f$ is an even polynomial with a positive leading coefficient? Do you want, for example, an $\varepsilon$-$N$ style proof? – Theo Bendit Apr 22 '21 at 05:02
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Yes, ε -N style proof would also be great! – Blake Apr 22 '21 at 05:06
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Note that \begin{align*} f(x) &= x^4 + x^3 + x^2 + x^2 + x + 1 \\ &= x^2(x^2 + x + 1) + x^2 + x + 1 \\ &= (x^2 + 1)(x^2 + x + 1). \end{align*} Further, note that $$x^2 + x + 1 = x^2 + x + \frac{1}{4} + \frac{3}{4} = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \ge \frac{3}{4},$$ and $x^2 + 1 > 0$, hence $$f(x) \ge \frac{3}{4}(x^2 + 1) \ge \frac{3}{4}x^2.$$ Now, this tells me that, given $M > 0$, we can choose $N = \frac{2\sqrt{M}}{\sqrt{3}}$, and get \begin{align*} |x| \ge N &\implies |x| \ge \frac{2\sqrt{M}}{\sqrt{3}} \\ &\implies x^2 \ge \frac{4M}{3} \\ &\implies \frac{3}{4}x^2 \ge M \\ &\implies f(x) \ge M. \end{align*} Thus, $f(x) \to \infty$ as $|x| \to \infty$.
Theo Bendit
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