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Given strongly convex $f(x)$ definded on $[a, b]$. I'm trying to prove that $\forall x \in (a, b)$ correct this strong inequality: $$f(x) < \max \{f(a), f(b) \}. $$ I can understand this fact on intuition level - from geometric side of convexity - every point under the chord, connecting any two points in interval, in which function defined. So each point is under the chord and under the $\max \{ f(a), f(b) \}$.

But how can i formalize it?

UPD: there was an misspell: i need it for strict, not strong convexity.

taciturno
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1 Answers1

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Recall that a function is strongly convex if and only if there is $m>0$ such that $f(y)\geq f(x)+\nabla f(x)(y-x)+\frac{m}{2}\lVert x-y \rVert^2$ for all $x,y$. Let $z= tx+(1-t)y$ for $0<t<1$, then: $$f(x)-f(z) \geq \nabla f(z)(x-z)+\frac{m}{2}\lVert x-z \rVert^2$$ $$f(y)-f(z) \geq \nabla f(z)(y-z)+\frac{m}{2}\lVert y-z \rVert^2$$ It follows that: $$\max\{f(x), f(y)\} \geq tf(x)+(1-t)f(y) \geq f(z)+ \frac{m}{2}\lVert x-z \rVert^2 +\frac{m}{2}\lVert y-z \rVert^2$$ Hence $f(tx+(1-t)y) < tf(x)+(1-t)f(y)\leq \max\{f(x), f(y)\}$. Note that we also showed that $f$ is strictly convex.

matiasdata
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    Can we use Jensen's Inequality here some way? – taciturno Apr 19 '20 at 09:11
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    And one more question: how we can say, that $\max {f(x), f(y) } \geq tf(x) + (1 - t) f(y)$? From what it follows? – taciturno Apr 19 '20 at 09:22
  • The inequality $ tf(x) + (1 - t) f(y)\leq \max {f(x), f(y) } $ follows since $ tf(x) + (1 - t) f(y)$ for $0\leq t \leq 1$ is the line segment from $f(y)$ to $f(x)$, hence they are all $\leq$ than the maximum of $f(x)$ and $f(y)$. This last inequality says that a convex function is quasi-convex. – matiasdata Apr 19 '20 at 12:48
  • there was a typo! I haven't seen it from start, but i need it for strict convexity more! Sorry. Do you know how to solve it for strict convexity? Maybe i could use Jensen's ineq, as i mentioned already? – taciturno Apr 19 '20 at 13:39
  • Strict convexity means that we have the inequality: $f(tx+(1-t)y) < tf(x)+(1-t)f(y)$ for $x\neq y$ and $0< t < 1$. But coupled with $tf(x) + (1 - t) f(y)\leq \max {f(x), f(y) }$, we get the desired strict inequality. – matiasdata Apr 19 '20 at 13:53