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I am dealing with a math problem which I can't resolve.

Given $f:\mathbb{R}\to\mathbb{R}$ twice differentiable with the following property:

$$ 2x f(x)\geq f'(0)-f'(x)$$

Show that

$$ f''(0)+f^2(0)\geq -1$$

I've been trying to use the derivative definition:

$$\lim \limits_{x \to x_0} \frac{f(x) -f(x_0)}{x-x_0}$$

But I got no result.

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  • $\begingroup$ Please use $ signs for inline math-formatting as well as for display; Use \geq for >=, and \leq for <=. Also, this is a math forum, and * is not used for multiplication. $\endgroup$ Commented Aug 22, 2018 at 13:34

3 Answers 3

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$f''(0)$ exists because $f$ is twice differentiable.

By definition $f''(0)=\lim\limits_{x\rightarrow0}\frac{f'(x)-f'(0)}{x}$

You can minorate the fraction using the given property $2xf(x)\geq f'(0)-f'(x)$ :

for $x>0$,$\frac{f'(x)-f'(0)}{x}\geq-2f(x)$

The LHS has a limit in $0$ because $f$ is twice differentiable. Taking the limits of continuous functions preserves inequalities, hence:

$f''(0)\geq-2f(0)$

From there

$f''(0)+f^2(0)\geq-2f(0)+f^2(0)=(1-f(0))^2-1$

And you can conclude using that a square is always non-negative.

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    $\begingroup$ @Evargalo thank you very much! $\endgroup$ Commented Aug 22, 2018 at 14:03
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hint

For $x>0,$

$$-2f(x)\le \frac{f'(x)-f'(0)}{x}$$

thus $$f''(0^+)\ge - 2f(0)$$

for $x<0$,

$$-2f(x)\ge \frac{f'(x)-f(0)}{x}$$

$$ \implies f''(0^-)\le -2f(0)$$

so $$f''(0)=-2f(0)$$

finally $$-2f(0)+f^2(0)+1=(f(0)-1)^2\ge 0$$

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Note that $$ f'(x)-f'(0)\ge-2xf(x) $$ and hence $$ \frac{f'(x)-f'(0)}{x}\ge-2f(x), \forall x>0 $$ and $$ \frac{f'(x)-f'(0)}{x}\le-2f(x), \forall x<0. $$ So $$ f''_+(0)=\lim_{x\to0^+}\frac{f'(x)-f'(0)}{x}\ge-2f(0) \tag{1}$$ and $$ f''_-(0)=\lim_{x\to0^-}\frac{f'(x)-f'(0)}{x}\le-2f(0). \tag{2}$$ From (1)(2), one has $$ f''(0)=-2f(0) $$ and hence $$ f''(0)+f^2(0)=-2f(0)+f^2(0)=(f(0)-1))^2-1\ge-1. $$

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