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I'm trying to prove that the function $f:\mathbb{R}\to \mathbb{R}$, $$ f(x) = |x|^p \quad ,\quad p \geq 1$$ is convex. By using the definition of a convex function and simplifying a bit, I arrived in the following inequality $$ |a+b|^p \leq |a|^p+|b|^p $$ where a and b are real numbers.

If I can prove that this holds, then $f$ is convex. However, I'm a bit lost here. I'm aware of the triangle inequality, but that only proves it for $p=1$. Is there such thing as a triangle inequality to the power of $p$? If not, any other suggestions would appreciated.

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    $\begingroup$ Let $p = 2$ and $x = y = 0.1$. Then $$|x+y|^p = |0.1 + 0.1|^2 = 0.2^2 = 0.04 > 0.02 = 0.01 + 0.01 = |0.1|^2 + |0.1|^2 = |x|^p + |y|^p$$ $\endgroup$ – TastyRomeo Nov 9 '16 at 10:24
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    $\begingroup$ Since you write $\lVert x\rVert^p$, I guess the domain of $f$ should be $\mathbb{R}^n$ rather than $\mathbb{R}$? Look at the function $g\colon [0,+\infty) \to [0,+\infty)$ given by $g(t) = t^p$. And think about the composition of convex functions. $\endgroup$ – Daniel Fischer Nov 9 '16 at 10:24
  • $\begingroup$ @SteamyRoot: thank you for the counter-example. However, it now turned the question moot. What's the best practice here, delete it? $\endgroup$ – JLagana Nov 9 '16 at 10:30
  • $\begingroup$ @DanielFischer: No, I actually meant $|x|$, instead of $||x||$. Thanks for pointing it out. $\endgroup$ – JLagana Nov 9 '16 at 10:31
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    $\begingroup$ Since $f$ is continuous, it suffices to prove midpoint convexity to deduce convexity. And that is $$\biggl\lvert \frac{a+b}{2}\biggr\rvert^p \leqslant \frac{\lvert a\rvert^p + \lvert b\rvert^p}{2}.$$ $\endgroup$ – Daniel Fischer Nov 9 '16 at 10:37
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No it does not hold for a,b greater than 1. Let b=ca. LHS is (1+c)^p a^p whereas RHS is (1+c)a^p . The reason it doesn't work is because x^p is increasingly increasing so it is more increasing between a and a+b than it is between 0 and b or 0 and a.

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