Show that $f(x)=||x||^p, p\ge 1$ is convex function on $\mathbb{R}^n$. I have tried to use Holder's inequality, but I still cannot solve this problem. Could you help me with this problem? Thank you so much.

  • 4
    $\begingroup$ This, I think, is one of those cases in which it is easier to prove the general case, rather than the particular one. Specifically, try to prove that the function $$F(u)=\Phi(f(u))$$ is convex whenever $f$ is convex and $\Phi$ is convex and nondecreasing. Then specialize it to your problem by setting $f(u)=\lVert u\rVert$ and $\Phi(v)=v^p,\ v\ge 0$. $\endgroup$ – Giuseppe Negro Oct 22 '13 at 17:01

This actually holds for any norm $||\cdot||$ on $\mathbb{R}^n$. Let $h:[0,\infty)\to[0,\infty)$ and $g:\mathbb{R}^n\to[0,\infty)$ be defined as

$$h(x):=x^p,\quad\quad g(x):=||x||.$$

where $p\geq 1$. From the triangle inequality we have that $g$ is convex. Also, it is easy to see that $h$ is non-decreasing and convex (since anywhere on $[0,\infty)$ the first and second derivatives of $h$ are non-negative).

So, for any $x,y\in\mathbb{R}^n$ and $\lambda\in [0,1]$,

$$f(\lambda x+(1-\lambda)y)=h(g(\lambda x+(1-\lambda)y))\leq h(\lambda g(x)+(1-\lambda)g(y))$$ $$\leq \lambda h(g(x))+(1-\lambda)h(g(y))=\lambda f(x)+(1-\lambda)f(y).$$

where the first inequality follows from convexity of $g$ and "increasingness" of $h$ and the second from convexity of $h$.

Edit: I just realised that the above was basically already stated in @GiuseppeNegro's comment. You can also generalise the above by substituting $\mathbb{R}^n$ with any convex normed vector space (over the reals).

  • $\begingroup$ @user52523 you're very welcome! $\endgroup$ – jkn Oct 22 '13 at 17:25

Assuming that $\|x\|$ is the Euclidean norm, we may solve it by showing that $f(t) = \|x+ty\|^p$ is convex for every choice of $x$ and $y$ in $\mathbb{R}^n$. Express $f:\mathbb{R}\rightarrow \mathbb{R}$ as follows: $$ f(t) = (x+ty,x+ty)^{p/2}, $$ where $(\cdot,\cdot)$ is the Euclidean inner product. $f$ is differentiable. It is convex if $f''(t)\geq 0$ for all $t$. Differentiating once, $$ f'(t) = \frac{p}{2} (x+ty,x+ty)^{p/2-1} 2(x+ty,y) . $$ Differentiating twice, $$ f''(t) = p(p/2-1)(x+ty,x+ty)^{p/2-2}(x+ty,y)^2 + p (x+ty,x+ty)^{p/2-1} (y,y). $$ Recognize terms as square norms to various powers: $$ f''(t) = p(p/2-1)\|x+ty\|^{p-1} (x+ty,y)^2 + p\|x+ty\|^{p-1/2}\|y\|^2 \geq 0. $$

  • 1
    $\begingroup$ It is given that $p\geq 1$. What happens when $1\leq p<2$? $\endgroup$ – daulomb Oct 22 '13 at 17:09
  • 1
    $\begingroup$ @user40615, if $p\in [1,2)$, just use Cauchy inquality in the expression. $(x+ty,y)^2$. $\endgroup$ – Tomás Oct 22 '13 at 17:14
  • 1
    $\begingroup$ @Simen there are some errors in the last line that you have to fix. It is $ f''(t) = p(p/2-1)\|x+ty\|^{p-4} (x+ty,y)^2 + p\|x+ty\|^{p-2}\|y\|^2 \geq 0. $ $\endgroup$ – Tomás Oct 22 '13 at 17:18
  • $\begingroup$ Thanks @Simen K.. I think that $$ f''(t) = p(p/2-1)\|x+ty\|^{p-1}2 (x+ty,y)^2 + p\|x+ty\|^{p-1/2}\|y\|^2 \geq 0. $$ $\endgroup$ – user52523 Oct 22 '13 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.