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I am looking for a example of a function in $C_0(\mathbb{R})$ such that $f',f'' \,\text{and}\, f''' \in C_0(\mathbb{R})$ with $$ \inf f < \inf (f-a*f''')$$ for some $a>0$, but I couldn't find one yet. I've tried functions like $$-\frac{1}{1+x^2}, -\frac{x^3}{1+x^4}, -e^{-x^2}$$, but none of these worked ... thank you for your answers in advance!

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Let $$f(x) = e^{-\frac{1}{x^2}}$$ and $a = 1$. Then $\inf f = 0$ and $$f'''(x) = \frac{e^{-\frac{1}{x^2}}}{x^{10}} \left( {8x-36x^3+24x^5} \right)$$ So $$f(x)-af'''(x) = \frac{e^{-\frac{1}{x^2}}}{x^{10}} \left( {-8x+36x^3-24x^5+x^{10}} \right)$$ $|f - a f'''|$ is finite (and indeed approaches $1$ for large $x$), and for sufficiently small positive $x$, is less than zero because of the leading $-8x$. So this satisfies your condition and is an answer.

If you are uncomfortable with the peculiar point at zero, where you have to assign $f(x) = \lim_{x \to 0} e^{-\frac{1}{x^2}}$ , then consider the function $$ f(x) = e^{-\frac{1}{1+x^2}}$$ where $\inf f = e^{-1}$. Again take $a=1$. Now $$f(x) - af'''(x) = \left( 1+12x-6x^2+40x^3+15x^4+12x^5+20x^6-24x^7+15x^8+6x^{10}+x^{12} \right)\frac{e^{-\frac{1}{1+x^2}}}{(1+x^2)^6} $$ which for small $x$ is $$ e^{-\frac{1}{1+x^2}} \left(1+12x -18x^2 + \ldots \right) $$ which when $x$ is a sufficiently small negative value is less than $e^{-1}$

Indeed, any even continuous 4-times differentiable function $f(x)$ with $\inf f$ occurring at $x=0$, having a bounded $f'''(x)$ and having $f''''(0) \neq 0$ will be a valid answer .

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  • $\begingroup$ Thank you for your effort! In the first example f-f''' takes negative values somewhere around -0.7 and 0.5 (see wolframalpha.com/input/?i=e%5E(−1%2Fx%5E2)%2F(x%5E10)(−8x%2B36x%5E3−24x%5E5%2Bx%5E10)) but it should be fine if one chooses a small enough. In the second example I don't know why you get this term, see wolframalpha.com/input/?i=d%5E3%2Fdx%5E3+exp(-1%2F(1%2Bx%5E2)). Anyway, I found f(x) = -(x+1)^2 *exp(-x^2) which has minimum approx. -1,786856 whereas f-0.001*f''' has minimum approx. -1,785314 > inf f. $\endgroup$
    – Stockfish
    Aug 20, 2016 at 22:58
  • $\begingroup$ But I don't think your last condition is enough. If f''' is strictly positive everywhere, then af''' is, too, and inf f will always be bigger than inf f-af'''. $\endgroup$
    – Stockfish
    Aug 20, 2016 at 23:08

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