6
$\begingroup$

Let $X$ be a vector space. I was able to prove that $\Vert\cdot \Vert:X\to \Bbb{R},$ is a convex function, i.e., for all $x,y\in X$ and $\lambda \in [0,1],$

\begin{align} \Vert \lambda x+(1-\lambda)y \Vert \leq \lambda \Vert x\Vert+(1-\lambda)\Vert y \Vert\end{align}

Now, I want to prove that $\Vert\cdot \Vert^2:X\to \Bbb{R},$ where $X$ is a vector space, is convex. So, here's what I've done!

MY WORK

\begin{align} \Vert \lambda x+(1-\lambda)y \Vert^2 \leq \left( \lambda \Vert x\Vert+(1-\lambda)\Vert y \Vert\right)^2,\;\;\text{for all}\;\; x,y\in X\;\; \text{and}\;\; \lambda \in [0,1].\end{align}

So, any help please on how to proceed?

$\endgroup$
  • 3
    $\begingroup$ do you want to specifically know if $\|\cdot\|^2$ is convex? then you should make this more clear, also in the title $\endgroup$ – supinf Nov 30 '18 at 10:40
  • $\begingroup$ @supinf: I made some edits! $\endgroup$ – Omojola Micheal Nov 30 '18 at 11:16
  • $\begingroup$ Does your vector space also provide inner product? $\endgroup$ – Mostafa Ayaz Nov 30 '18 at 11:54
2
$\begingroup$

In general if $f$ is a convex function and $g$ is a convex nondecreasing function then the composition $g \circ f$ is a convex function. Let $f(\cdot)=\|\cdot \|$ which maps to $\mathbb{R}_{\geq 0}$ and let $g(x)=x^2$ which is a nondecreasing convex function on $\mathbb{R}_{\geq 0}$. If follows that $g \circ f (\cdot)=\| \cdot \|^2$ is a convex function.

See The composition of two convex functions is convex for the original claim.

$\endgroup$
  • $\begingroup$ This is good! I like it! $\endgroup$ – Omojola Micheal Nov 30 '18 at 11:31
4
$\begingroup$

Just solved and thought to share it for the sake of future readers. \begin{align} \Vert \lambda x+(1-\lambda)y \Vert^2 &\leq \left( \lambda \Vert x\Vert+(1-\lambda)\Vert y \Vert\right)^2\\ &\leq \lambda^2 \Vert x\Vert^2+2\lambda(1-\lambda)\Vert x\Vert\Vert y\Vert+ (1-\lambda)^2\Vert y\Vert^2\\ &= \lambda^2 \Vert x\Vert^2+2\lambda(1-\lambda)\Vert x\Vert\Vert y\Vert+ (1-\lambda)^2\Vert y\Vert^2 -\lambda\Vert x\Vert^2 -(1-\lambda)\Vert y\Vert^2\\&\quad+\lambda\Vert x\Vert^2 +(1-\lambda)\Vert y\Vert^2,\;\;\text{adding and substracting}\;\lambda\Vert x\Vert^2 +(1-\lambda)\Vert y\Vert^2 \\ &= -\lambda (1-\lambda)\left(\Vert x\Vert-\Vert y\Vert\right)^2+\lambda\Vert x\Vert^2 +(1-\lambda)\Vert y\Vert^2\\ &\leq \lambda\Vert x\Vert^2 +(1-\lambda)\Vert y\Vert^2,\;\;\text{since}\;-\lambda (1-\lambda)\left(\Vert x\Vert-\Vert y\Vert\right)^2\leq 0.\end{align} Hence, $\Vert\cdot\Vert^2$ is a convex function.

$\endgroup$
  • $\begingroup$ Nice! (+1)...... $\endgroup$ – Mostafa Ayaz Nov 30 '18 at 11:53
1
$\begingroup$

Define $p=\lambda x$ and $q=(1-\lambda)y$, therefore we need to show that $$||p+q||^2\le (||p||+||q||)^2$$which reduces to $$p\cdot q\le ||p||\cdot ||q||$$which is the same famous Cauchy-Schwartz inequality. Therefore $||.||^2$ is convex.

$\endgroup$
  • $\begingroup$ That's fine too! $\endgroup$ – Omojola Micheal Nov 30 '18 at 11:31
  • $\begingroup$ Thank you. Good luck! $\endgroup$ – Mostafa Ayaz Nov 30 '18 at 11:33
  • $\begingroup$ This only works if the norm comes from a scalar product, however. $\endgroup$ – Giuseppe Negro Nov 30 '18 at 11:50

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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