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Let $f:\mathbb{R}\to\mathbb{R}$ be a convex function.

How can I prove that for each $x$, there is $c$ such that $f(x)+c(y-x)\leq f(y)$ for all $y$?

One of the difficulties to solve is $f$ does not need to be differentiable. It makes me feel hard. So I must show this inequality by only using the definition of a convex function: $$f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$$ holds for any $\lambda\in[0,1]$.

Analysis
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1 Answers1

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Hint:

Take $x$ real and $t\neq 0$. Define $$v(t)=\frac{f(x)-f(x+t)}{t}$$

Show that $v$ is increasing. Define $$c^-=\lim_{t\to0^-} v(t)$$ and $$c^+=\lim_{t\to0^+}v(t)$$

Now, take $c\in[c^-,c^+]$.

ajotatxe
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