Sufficient condition for convexity

Question

Let f:$ [a,b] \rightarrow \mathbb{R} $ a continous function

such that $ \forall (x,y) \in [a,b]^{2}, \exists t \in ]0,1[, f(tx+(1-t)y) \le tf(x) + (1-t)f(y) $

show that f is convex

Answer 1

The idea is that a convex combination of convex combinations of two numbers $a$ and $b$ is itself a convex combination of $a$ and $b$,

Assume by contradiction that $f$ is not convex. There exist $x_0,x_1\in [a,b],\ t_0\in[0,1]$, ($x_0<x_1$ for simplicity) such that $f(x_{t_0})>(1-t_0)f(x_0)+t_0f(x_1)$, where $x_{t}:=(1-t) x_0+t x_1$, $0\le t\le 1$.

You prove that $0<t_0<1$.

Denote by $g(t):=f(x_{t})-(1-t)f(x_0)-tf(x_1)$. Since $g$ is continuous and positive at $t_0\in(0,1)$ it stays positive on a whole non-degenerate interval $[u,v]\subset(0,1)$.

Let $\alpha:=\inf \{t\in[0,\alpha]\mid g(s)>0.\forall\ s\in[t,v]\}$.

You prove that $g(\alpha)=0$. Also, you need to define similarly $\omega>\alpha$ such that $g(\omega)=0$ and $g$ stays positive on $(\alpha,\omega)$.

Now we use the given assumption for $x=x_\alpha$, $y=x_\omega$ to get a $t\in(0,1)$ such that $f(tx_\alpha+(1-t)x_\omega)\le tf(x_\alpha)+(1-t)f(x_\omega)$ together with the facts that $tx_\alpha+(1-t)x_\omega=x_s$ (for what $s$? -- very easy to guess), $s\in(\alpha,\omega)$, $f(x_\alpha)=(1-\alpha)f(x_0)+\alpha f(x_1)$, and similarly for $f(x_\omega)$. You get a contradiction from the previous inequality. You need to complete all details and/or simplify this proof.

Where is the condition $t\in(0,1)$ essential?

Answer 2

A proof might go like this (i.e., this is incomplete):

Suppose that $f$ is not convex. Then there are two points $P=(u, f(u))$ and $Q=(v, f(v))$ such that the line between them is sometimes below $f$; i.e., there is a $t \in (0, 1)$ such that, if $w = tu+(1-t)v$, then $f(w) < tf(u)+(1-t)f(v)$.

Since $f$ is continuous, there is a region around $w$ where the line connecting such points is completely below $f$. This is the part I'm not completely sure of how to prove. I think that we could narrow regions where a point on the line is below $f$ and use the continuity of $f$ to establish an interval where the line is below the curve.

Answer 3

Suppose $f$ is not convex. It means there is an interval $[x,y]\subset[a,b]$ where the line between $x$ and $y$ is above the curve: $$ \exists x,y\in[a,b],\forall t\in[0,1]:f(tx+(1-t)y) > tf(x) + (1-t)f(y) $$ which is equivalent to $$ \exists x,y\in[a,b],\not\exists t\in[0,1]:f(tx+(1-t)y) \le tf(x) + (1-t)f(y) $$ This contradicts the hypothesis.

Answer 4

By bissection, you can go to $\forall(x,y)\in[a,b]^2,\forall t\in]0,1[, f(tx+(1−t)y)≤tf(x)+(1−t)f(y)$

Let $(x,y) \in [a,b]^2$, let $t^* \in [0,1]$. We want to prove that $$f(t^*x+(1−t^*)y)≤t^*f(x)+(1−t^*)f(y)$$. By property, there exist $t_0$ such as

$$f(t_0x+(1−t_0)y)≤t_0f(x)+(1−t_0)f(y)$$

Let $x_0 = t_0x + (1 -t_0)y$. If $t \in [x, x_0]$, there exist $t_1$ such as

\begin{eqnarray} f(t_1x+(1−t_1)x_0)) & ≤ & t_1f(x)+(1−t_1)f(x_0) \\ & \leq & t_1f(x)+(1−t_1)t_0f(x)+(1−t_0)f(y) \\ & \leq & (t_1 + (1-t_1)t_0) f(x) + (1-t_1)(1-t_0)f(y) \end{eqnarray}

and so forth, you have $t \in [t_{i} , t_{j}]$, with inequality verified by $x_i$ and $x_j$ and both converging to $t$. With continuity of $f$, it's done.