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If $f: R^n\rightarrow R$ is convex and $f(\alpha x)=\alpha f(x), \alpha \geq 0$, show that:
a) $f(x+y)\leq f(x)+f(y)$ for all $x,y\in R^n$
b) $f(0)\geq 0$
c) $f(-x)\geq -f(x)$ for all $x\in R^n$
d) $f(\alpha_1 x_1 +...+\alpha_m x_m)\leq \alpha_1f(x_1)+...+\alpha_mf(x_m)$ for all $\alpha_k>0, x_k\in R^n, k=1, .., m$

Can you help me to prove these inequalities? I apologize for writing them all in one question, but I can't write more questions here today and I need solutions for tomorrow morning.

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  • $\begingroup$ What are you having difficulty with? $\endgroup$
    – copper.hat
    Jun 11, 2013 at 16:37
  • $\begingroup$ @Landscape I corrected b). Thanks. $\endgroup$
    – user23709
    Jun 11, 2013 at 16:59
  • $\begingroup$ @user23709: You are welcome. Please note that in fact $f(0)=0$(let $\alpha=0$). $\endgroup$
    – 23rd
    Jun 11, 2013 at 17:02
  • $\begingroup$ @copper.hat I now have problem with a) and d), since I got answer for b) and c). I don't know what to do with that. $\endgroup$
    – user23709
    Jun 11, 2013 at 17:04
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    $\begingroup$ For a) (and d is similar), note that $x+y = 2 (\frac{1}{2}x + \frac{1}{2} y)$. Use convexity to obtain an upper bound on $f(\frac{1}{2}x + \frac{1}{2} y)$ and positive homogeneity to finish the proof. $\endgroup$
    – copper.hat
    Jun 11, 2013 at 17:11

1 Answer 1

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If your function is continuous. We have for $t \in (0,1)$ \begin{equation} f(tx+(1-t)x) \le tf(x) + (1-t)f(y) \le tf(x) +f(y) \end{equation} Doing $t $ tends to one we obtain a).

To b) $f(0) = f(0x) =0f(x) =0$.

to c) $0 = f(x-x) \le f(x) +f(-x)$

To d) a do simple indction of a)

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  • $\begingroup$ So, in c) you used part a) ? $\endgroup$
    – user23709
    Jun 11, 2013 at 17:00
  • $\begingroup$ You don't need to invoke continuity here, and the above does not show subadditivity. It just shows $f(x) \le f(x)+f(y)$ (assuming the second $x$ was meant to be $y$). (Any convex function defined on all of $\mathbb{R}^n$ is continuous, but this is not required here.) $\endgroup$
    – copper.hat
    Jun 11, 2013 at 17:15

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