Prove that $\Vert\cdot \Vert^2:X\to \Bbb{R},$ where $X$ is a vector space, is convex

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?

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

• This is good! I like it! – 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.

• Nice! (+1)...... – 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.

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