Suppose that $f$ is not strictly convex on $C$. Prove that there exist $x ,y \in !n(x \not= y)$ such that $f$ is affine over the segment $[x, y]$

Question

Let $f$ be a convex function defined on a convex set $C$. Suppose that $f$ is not strictly convex on $C$. Prove that there exist $x, y \in \mathbb{R}^n,(x \not= y)$ such that $f$ is affine over the segment $[x, y]$.

My idea was assuming by negativity that, for every $x, y \in \mathbb{R}^n$, $f$ is not affine and therefore because $f$ is convex it must be strictly convex.

Sttart with the definition of “not strictly convex”. Then this should help: A convex function which intersects a chord at three points is affine. — Martin R, Nov 10 '20 at 12:18
@MartinR $f(\lambda x+(1-\lambda)y)<\lambda f(x)+(1-\lambda)f(y)$ — convxy, Nov 10 '20 at 12:23
“Convex, but not strictly convex” means that $f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)$ for some $x, y \in C$ and some $\lambda \in (0, 1)$. Now use the other question to show that $f$ is affine on $[x, y]$. — Martin R, Nov 10 '20 at 12:25
@MartinR I've thought about it and the third proof in the link is not valid for R^n how can i fix it? — convxy, Nov 13 '20 at 13:32
You can reduce it to a one-dimensional problem by considering $g(t) = f(x + t(y-x))$ – as in your other recent question. — Martin R, Nov 13 '20 at 14:16

Maverick Meerkat · Answer 1 · 2021-02-05T11:34:33.290

Since $f$ is not strictly convex, $\exists x,y\in \mathbb R^n, \lambda\in(0,1)$ s.t. $z =\lambda x+(1-\lambda)y, \; f(z)=\lambda f(x)+(1-\lambda)f(y)$.

Claim: $f$ is affine in $[x,y]$, i.e. for any point $p\in[x,y], p=\lambda'x+(1-\lambda')y, f(p)=\lambda'f(x)+(1-\lambda')f(y)$.

Proof: Suppose towards contradiction that it isn't affine. I.e. there exists one point, $u\in (x,y)$ s.t. $f(u)\neq\lambda'f(x)+(1-\lambda')f(y)$. By the convexity of $f$ it must be that $f(u)<\lambda'f(x)+(1-\lambda')f(y)$. We know that the point cannot be $x,y$, or $z$, so it must either be in $(x,z)$ or in $(z,y)$. Assume WLOG that the point is $\in(z,y)$ and denote it as a convex combination using $\alpha \in (0,1)$, s.t. $$u=\alpha z + (1-\alpha)y$$ $$f(u)=f(\alpha z+(1-\alpha)y)<\alpha f(z)+(1-\alpha)f(y)=\alpha\lambda f(x)+(1-\alpha\lambda)f(y)$$ (Note, we can do this because $x=\frac{z-(1-\lambda)y}{\lambda}, \; f(x) = \frac{f(z)-(1-\lambda)f(y)}{\lambda}$ hence $u = \frac{\lambda'}{\lambda}z +(1-\frac{\lambda'}{\lambda})y$ so we just denote $\alpha = \frac{\lambda'}{\lambda}$. $\lambda'f(x)+(1-\lambda')f(y)=\lambda'\frac{f(z)-(1-\lambda)f(y)}{\lambda} + (1-\lambda')f(y)=\alpha f(z)+(1-\alpha)f(y)$)

Now, let us choose symmetrically the point $v = \alpha x + (1-\alpha)z$. From convexity we get that $$f(v) \le (\alpha + \lambda -\alpha\lambda)f(x) + (1-\alpha-\lambda -\alpha\lambda)f(y) $$ Let us choose $$ t = \frac{\alpha\lambda-\alpha}{2\alpha\lambda-\alpha-\lambda}$$ It's easy to show that (for $\alpha,\lambda\in(0,1)$) $t\in(0,1)$. First $\alpha\lambda<\alpha$ and $\alpha\lambda < \lambda$ - so both the numerator and the denominator are negative, hence $t$ is positive. Second, the denominator is the numerator + another negative quantity: $\alpha\lambda-\alpha + \alpha\lambda - \lambda$ hence it's magnitude is bigger.

Note that $z=tu + (1-t)v$. By convexity we get that $f(z)=f(tu + (1-t)v) \le tf(u)+(1-t)f(v)$. By the strict inequality of $f(u)$ we get that $$< t(\alpha\lambda f(x)+(1-\alpha\lambda)f(y)) +(1-t)((\alpha + \lambda -\alpha\lambda)f(x) + (1-\alpha-\lambda -\alpha\lambda)f(y))=\lambda f(x) + (1-\lambda)f(y) $$ Which is a contradiction to the assumption that $f(z) = \lambda f(x) + (1-\lambda)f(y)$.

Hence $f$ is affine in $[x,y]$.

Suppose that $f$ is not strictly convex on $C$. Prove that there exist $x ,y \in !n(x \not= y)$ such that $f$ is affine over the segment $[x, y]$

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