I was proving the following question:
Prove that if $u:\mathbb{R}\to \mathbb{R}$ is twice differenciable, so:
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$u$ is concave.
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$u'' \leqslant 0$.
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$u'$ is decresing.
I had no problem to prove it with the following definition of concave function:
$$ u(x)\leqslant u(c) + u'(c)(x-c) $$
The problem was that I wanted to prove it with the following definition of concave function:
$$ Let \; x,y \in \mathbb{R}, \; so \; f(\alpha x + (1-\alpha) y) \geqslant \alpha f(x) + (1- \alpha) f(y) , \; \forall \alpha \in [0,1] $$
But I couldn't prove the equivalence of this two definitions in when we are talking about defferenciable functions.
Can someone help me with this equivalence?
Best Answer
Take a look at the proof of theorem 6 [[here]][1] but set $n=1$. But you don't need the full equivalence, just one way. There's a simpler statement out there for a single variable concave function, sometimes called the "rooftop theorem" (if $f$ is a single variable function that is concave and differentiable then that first inequality with the first order Taylor approximation holds....)
Basically, if $\alpha \in (0,1)$
$$ f(\alpha x + (1 - \alpha)y) \geq \alpha f(x) + (1 - \alpha)f(y)$$ $$f(\alpha (x-y) + y) \geq \alpha (f(x) - f(y)) + f(y) $$ $$f(x) - f(y) \leq \frac{f(\alpha(x-y) + y) - f(y)}{\alpha}$$ $$f(x) \leq \frac{f(\alpha(x-y) + y) - f(y)}{\alpha} + f(y)$$ $$f(x) \leq \frac{f(\alpha(x-y) + y) - f(y)}{\alpha(x-y)}(x-y) + f(y)$$
and take the limit when $\alpha \to 0$, to get $f'(y)$.