# Is $L_2$-norm a strictly convex function?

I am new to convex analysis, and just wondering whether there is a simple check to see whether $L_2$-norm is strictly convex. How to mathematically prove/disprove this?

$L_2$-norm: $\| x\|_2 = \sqrt{\sum x_i^2}$.

• When $n = 1$, the $L_2$ norm is just the absolute value function, which you can see clearly is not strictly convex. (The picture is also clear when $n = 2$, and the graph of the $L_2$ norm looks like an ice cream cone.) – littleO Jul 10 '15 at 21:20
• Do you mean strictly convex in the sense that you have $$\lVert \lambda x + (1-\lambda)y\rVert < \lambda \lVert x\rVert + (1-\lambda)\lVert y\rVert$$ for all $x\neq y$ and $0 < \lambda < 1$, or in the sense "strictly convex norm", i.e. that the unit sphere contains no line segment? If the latter, note that norms arising from inner products are even uniformly convex. – Daniel Fischer Jul 10 '15 at 21:22
• @DanielFischer, I mean the first one. – Aaron Zeng Jul 10 '15 at 21:25
• @littleO, is there formal mathematical proof on this? – Aaron Zeng Jul 10 '15 at 21:26
• @A.G.'s answer is a formal mathematical proof. All you have to do is provide an example to show that the definition of strict convexity is not satisfied. A simple example is to let $x = 0$, $y = (1,0,\ldots,0)$, and $\lambda = 1/2$. – littleO Jul 10 '15 at 21:46

This answer gives a reply to the question if a space equipped with $L_2$ norm is a strictly convex space. The answer to which is "yes." This however appears not be what was asked for. I still leave this up for now as I think it might help to avoid potential confusions.

Intuitively, strictly convex means that if you have two points on the boundary on the unit ball then the line segment connecting these two points is in the interior of the unit ball (and not at the boundary).

For the $L_2$ norm (say in $\mathbb{R}^3$ where we can visualize things) the unit ball is an "actual" ball, so if you have two points on the surface an you connect them with a straight line then the line goes though the interior of the ball.

Contrast this with the $L_{\infty}$ norm where the unit ball has the form of a cube. When you take two points on the same face and connect them you stay on the face that is on the boundary.

Rigorously, it is a quite direct consequence of the characterization of the case of equality in the triangle inequality which in turn rest on the characterization of equality in the Cauchy-Schwarz inequality.

What you need to show is that for distinct $x,y$ with $\| x \| = \| y \| = 1$ and for $0 < \lambda < 1$ you have $$\| \lambda x + (1 - \lambda ) y \| < 1.$$ Note that $\lambda x + (1 - \lambda ) y$ is a point on the line connecting $x$ and $y$. Further note that by the triangle inequality and the linearity of the norm you have right away $$\| \lambda x + (1 - \lambda ) y \| \le \| \lambda x \| + \|(1 - \lambda ) y \| = \lambda \| x\| + (1 - \lambda) \| y \| = \lambda + (1 - \lambda) = 1.$$ Thus, if the triangle inequality is strict then you would have what you want.

Now, recall that you have equality in the triangle inequality if and only if $x = \mu y$ or $y = \mu x$ for a positive $\mu$. However, since $x$ and $y$ both have norm $1$ and are distinct this can never happen.

To see this fact on the triangle inequality recall that for equality you need $\langle x , y \rangle = \|x \| \| y \|$, which only holds under that condition.

This arguments rests true for any norm that comes from an inner product.

• "Strictly convex" (as some other convexity related notions) differs for sets and for functions. In this sence, a unit ball is a strictly convex set, however, a norm as a function is not strictly convex. "Convex norms" in topology is a slang. – A.Γ. Jul 10 '15 at 21:47
• Yes, thanks, I know (as might be visible from my comments on OP). I was in the process of adding a remark at the start. – quid Jul 10 '15 at 21:48

For $x$ and $y=2x\ne x$ we have for any $\lambda\in(0,1)$ $$\|\lambda x+(1-\lambda)y\|=\|(\lambda+(1-\lambda)2)x\|=\lambda \|x\|+(1-\lambda)2\|x\|=\lambda \|x\|+(1-\lambda)\|y\|$$ which means no, $f(x)=\|x\|$ is not a strictly convex function. If you think of the function graph (the cone) you will see that it becomes flat on the radial rays, i.e. the Hessian (second derivative) degenerates. However, the function $\|x\|^2$ is strictly convex.

• For norms on a vector space, "strictly convex" has a different meaning, namely that the boundary of the unit ball does not contain any line segment. It's not clear which meaning the OP intended. – Daniel Fischer Jul 10 '15 at 21:19
• That's a good example. But is there a formal mathematical proof, or any related theorem to this. – Aaron Zeng Jul 10 '15 at 21:26
• @DanielFischer Okay, I have to add that I meant $f(x)=\|x\|$ is not strictly convex function. – A.Γ. Jul 10 '15 at 21:26
• Turns out that was what the OP intended. All is jolly. – Daniel Fischer Jul 10 '15 at 21:28
• @Aaron Zeng By definition of a strictly convex function $$f(\lambda x+(1-\lambda)y)<\lambda f(x)+(1-\lambda)f(y),\ \forall x\ne y,\ \forall \lambda\in(0,1)$$ thus to disprove the strict convexity you have to find some $x$ and $y\ne x$ and some $\lambda\in(0,1)$ which give equality (it is called a counterexample). My $y=2x\ne x$ above is a counterexample that works for any $x\ne 0$ and for any $\lambda\in(0,1)$. – A.Γ. Jul 10 '15 at 21:39