Distance between the minimum of two strongly convex functions and the minimum of their sum

257 Views Asked by At

Let $f$ and $g$ be two strongly convex functions from a convex set $\mathcal{X}$ to $\mathbb{R}$, with minimum $x$ and $y$ respectively. Let us denote by $z$ the minimum of $f+g$, and by $\delta$ the distance $\lVert x-y\rVert$.

I would like to know if it is possible to prove that $\lVert x-z\rVert \leq \delta$ and $\lVert y-z\rVert < \delta$.

Rk:

  • This statement is true in one dimension (see here)

  • This statement is also true when $f$ and $g$ are two quadratic functions.

1

There are 1 best solutions below

3
On

This result is true at least for any $p$-norm with $p\geq 1$. Indeed, we know that the result is true in one dimension (see here). Thus, the result is still true if we apply it to each of the coordinates of the functions $f$ and $g$.

In particular, we get: $$ \forall 1 \leq i \leq n,\ \min(x_i, y_i) \leq z_i \leq \max(x_i, y_i) $$

This shows us that $$ \lVert x-z \rVert_p = \left(\sum_{i=1}^n (x_i-z_i)^p\right)^{1/p} \leq \left(\sum_{i=1}^n (x_i-y_i)^p\right)^{1/p} = \lVert x_i - y_i \rVert_p =\delta $$ By symmetry, the same result holds for $y$: $\lVert y-z \rVert_p \leq \delta$.