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Is it possible to find a differentiable $f:\mathbb{R} \rightarrow \mathbb{R}$ such that

$$ xf(x) \geq L \quad \forall x \in \mathbb{R} $$

for an arbitrary constant $L>0$ ?

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up vote 1 down vote accepted

Hint. See what happens for $x = 0$.

At every other point $f(x) = L/x$ would work.

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exactly thats my problem... Can I find $f$ such that it "sucks" this zero for example with $0*1/0$ yielding a strictly positive number and $f$ be differentiable? – niels Sep 26 '12 at 7:53
No, $0y = 0$, there is nothing you can do about it. – Karolis Juodelė Sep 26 '12 at 7:54

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