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prove that $$ f(x,t)=\frac{\Vert x\Vert _p^p} {t^{p-1}} $$ is convex on $$ \{ (x,t)|x\in R^n ,t\ge 0 \} $$

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  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Oct 31 '17 at 14:25
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It is non-trivial to test definite of the Hessian matrix. As we know, there are many ways to prove convexity of the function. $$f\left( {x,t} \right) = \frac{{\left\| x \right\|_p^p}}{{{t^{p - 1}}}} $$ is the perspective function of $$ f\left( {x} \right)={\left\| x \right\|_p^p}$$ As we know, if perspective of a function is convex, then the function is also the convex function. You can prove the f(x) is a convex function. $$ f\left( {x} \right)={\left\| x \right\|_p^p}={\sum\limits_{i = 1}^n {\left| {{x_i}} \right|} ^p}$$ Then the f(x) is obvious a funcion by check its Hessian matrix (every norm on R^n is convex due to the triangel inequality of norm)

Another way to prove, you may use the Holder's inequality or by the epigraph of the function f(x,t).

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  • $\begingroup$ why the perspective of f(x) is convex? can you show more detail on norm function proof? $\endgroup$ – Vicky Oct 31 '17 at 15:51

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