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!


\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?

  • 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

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.

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

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.

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

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.

  • $\begingroup$ 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

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