If $f(y)=\limsup_k\|y-x_k\|^2$ then $f$ is a strictly convex function

Question

Let $y \in \mathbb{R^n}$, and $(x_k)_{k\in \mathbb{N}}$ a sequence in $ \mathbb{R^n}$ and $f(y)=\limsup_k\|y-x_k\|^2$ ($\|{\cdot}\|$ is the euclidean norm).

Then $f$ is a convex function.

But how can we prove that $f$ is a strictly convex function ?

Answer 1

I'm assuming that $(x_k)$ is a bounded sequence, so that $f(y)$ is finite for all $y \in \Bbb R$.

Generally, for $a, b \in \Bbb R^n$ and $\lambda \in \Bbb R$ we have, with $\cdot$ denoting the scalar product: $$ \Vert \lambda a + (1-\lambda) b \Vert^2 = \lambda^2 \Vert a \Vert^2 + 2 \lambda(1-\lambda) (a \cdot b)+ \lambda^2 \Vert b \Vert^2 \\ = \lambda \Vert a \Vert^2 + \lambda \Vert b \Vert^2 - \lambda (1-\lambda)\left( \Vert a \Vert^2 - 2 (a \cdot b)+ \Vert b \Vert^2 \right)\\ = \lambda \Vert a \Vert^2 + \lambda \Vert b \Vert^2 - \lambda (1-\lambda) \Vert a-b \Vert^2 \, . $$

Now let $y, z \in \Bbb R^n$, $y \ne z$, and $0<\lambda< 1$. For all $k$ we have $$ \Vert \lambda y + (1-\lambda) z - x_k \Vert^2 = \Vert \lambda(y-x_k) + (1-\lambda) (z-x_k) \Vert^2 \\ = \lambda \Vert y-x_k \Vert^2 + (1-\lambda) \Vert z-x_k \Vert^2 - \lambda (1-\lambda) \Vert y-z \Vert^2 \, . $$ It follows that $$ f(\lambda y + (1-\lambda) z) \le \lambda f(y) + (1-\lambda) f(z) - \lambda (1-\lambda) \Vert y-z \Vert^2 \\ < \lambda f(y) + (1-\lambda) f(z)\, . $$

Answer 2

we first prove that $g_k(x)=||x-x_k||^2$ is a convex function.

Then we will have $g_k(tx+(1-t)y)\leq tg_k(x)+(1-t)g_k(y)$ for all $t\in [0,1] $ and $x,y\in \mathbb{R^n}.$

which means $$||tx+(1-t)y-x_k|| \leq ||x-x_k||+ ||y-x_k||$$

then $$limsup_k||tx+(1-t)y-x_k|| \leq limsup_k||x-x_k||+ limsup_k||y-x_k||$$ Hence, $f$ is convex.