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Assume $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function and for some real number $a$ and all real numbers $x$, $$ f(x)+(x-a)f'(x)\gt0 $$ Prove that $f(x)=0$ has no real root. I tried to show $g(x)=xf(x)$ has only one root.. I calculated $g$'s derivative: $$ g'(x)=f(x)+xf'(x) $$ Then I tried to use the mean value theorem to show that $g$ is one-to-one. I concluded if there are $x_1$ and $x_2$ such that $x_1\lt x_2$ and $f(x_0)=f(x_2)$, then there is a $x_3$ such that $g(x_3)=0$ and $$ 0 \gt af(x_3) $$ But I don't know where I should go now.

3 Answers3

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This statement is incorrect the way it is given: $f(x)=0$ is a trivial counterexample.

However, if we "fix" the conditions to state $f(x)+(x-a)f'(x)\color{red}{\gt} 0$ (rather than $\color{red}{\ge} 0$), then the statement can be proven. Note:

$$\frac{d}{dx}((x-a)f(x))=f(x)+(x-a)f'(x)\gt 0$$

so $(x-a)f(x)$ is a (strictly) increasing function. This function has one zero (at $x=a$) and so $(x-a)f(x)\ne 0$, implying $f(x)\ne 0$, for $x\ne a$.

In addition, if we substitute $x=a$ in the above "fixed" condition, we get:

$$f(a)+(a-a)f'(a)\gt 0$$

so $f(a)\ne 0$. In other words, $f$ has no zeros, neither equal to or different from $a$.

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Suppose $f(x)+(x-a)f'(x)>0$, $f(a)>0$ write $g(x)=(x-a)f(x)$, $g'(x)>0$ implies $x>a, g(x)=(x-a)f(x)>g(a)=0$ implies $f(x)>0$

$x<a, g(x)<g(a)=0$ implies that $(x-a)f(x)<0$, $f(x)>0$

Tsemo Aristide
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HINT.-Clearly $x\notin\{0,a\}$and the condition is equivalent to $\dfrac{f'(x)}{f(x)}\ge\dfrac{-1}{x-a}$; we have taking integrals $$\log|f(x)|\ge-\log(x-a)$$ This evidently implies that $f (x) = 0$ would imply an absurdity ($-\infty\ge b\in\mathbb R)$

Piquito
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  • Just writing $f'(x)/f(x)$ does not already implies $f(x)\neq0$? – Vincenzo Tibullo Aug 08 '20 at 15:33
  • True, but that would need for the proof of an additional calculation that seems to me less forceful than minus infinity greater or equal than a real number. I'm right? – Piquito Aug 08 '20 at 17:13