I want to prove that if function $f:(a,b)\to \mathbb{R}$ is not constant, increasing and convex then from some $c\in (a,b)$ function $f:[c,b)\to \mathbb{R}$ is strictly increasing. Any ideas?
[Math] Convex function strictly increasing
calculusreal-analysis
Related Solutions
One property that will give us $f'$ strictly increasing is the following : we say that $f$ is convex if $$ \forall \lambda \in [0,1], \quad f(x + \lambda( y-x)) \le f(x) + \lambda( f(y) - f(x) ) $$ and we say that $f$ is strictly convex if the inequality is always strict. (The intuition behind this is that if you draw the cord between the points $(x,f(x))$ and $(y,f(y))$, the cord cannot touch the graph, thus we cannot have an interval where the derivative is constant if $f$ is strictly increasing.)
Assume $f$ is strictly convex and differentiable. Then $f'$ is strictly increasing. Since strictly convex functions are convex, we already know that $f'$ is increasing. Suppose there exists $x \le y$ such that $f'(x) = f'(y)$. Then this means that in the interval $[x,y]$, the function $f'$ is constant. But then $f$ cannot be strictly increasing on this interval unless $x=y$, thus $f'$ must be strictly increasing. The reason for this is that strict convexity on $[x,y]$ with $x < y$ implies that by the mean value theorem, there exists $c_1$, $c_2 \in ]x,y[$ with the property that $$ \begin{align} f(x + \lambda (y-x)) - f(x) & < \lambda (f(y) - f(x)) \\ \\ \Longrightarrow \quad f'(c_1) = \frac{ f(x + \lambda(y-x)) - f(x)}{\lambda(y-x)} & < \frac{f(y) - f(x)}{y-x} = f'(c_2) \end{align} $$ which is a contradiction. One way to ensure that a function has a strictly decreasing derivative is strict concavity : a function is strictly concave if equality is reversed in the definition of strict convexity and the proof is similar. Is this what you were looking for? I hope it helps.
hmmmmm. I am not sure. The property you mention is not generally a property (in general) of either concave or convex functions, but actually of sub-additive functions. However, I am inclined to believe that the conjecture, or the book may be correct for the following reasoning.
Technically, (on the average, to abuse terminology) a property that would most likely be shared by concave functions over the positive reals, rather then convex functions. Unless its one of these sub-linear or super-linear functions which are some admixture of both (and almost, both,'convex and concave', or super-additive, and sub-additive) over $[0,1]$)
Notice that in $(1)$ the scope of the multiplicative elements $a$ and $(1-a)$ is within $F$ not outside $F$ as in $(2)$ and $(3)$. If one can show that $F$ is sub-additive or that F(tx)>=tF(x) it should be.
$(1)f(ax+(1−a)y)≤f(ax)+f((1−a)y)$
Convex: $f(ax+(1−a)y) ≤a f(x)+(1−a)f(y)$
Concave: $f(ax+(1−a)y)\geq af(x)+(1−a)f(y)$
I agree that $(1)$ definitely Seems wrong for concave $F$ but it actually may not be. Remember that the "above" $(1)$ is a property that function, $F$ 'sub-additive over the positive reals' may have.
Concave functions are sub-additive, over the positive reals, when $f(0) ≥ 0$,see https://en.wikipedia.org/wiki/Concave_function. See pt 10. Although the domain only specifies positive- reals, not-non-negative reals and may not include the $0$. That is the only the issue. It also not a closed domain and range either.
However, given the positive domain, and strict monotonic increasing-ness, it might have the same effect, and the conjecture MIGHT (I stress) be correct. It will also be strictly quasi convex and strictly quasi concave.
If $F$ is concave and,$f(0) ≥ 0$ then $F$ is sub-additive over the positive reals; so the book, is "arguably" correct.
I (stress) arguably, as the the domain, of the function, only specifies positive- reals, not-negative reals. ie including $0$. Maybe strict mono-tonicity, and posi-tivity may help in this case. Notice, that given sub-additivity.
Best Answer
Lemma. For $\forall \,u \lt v \lt w \in (a,b)\,$:
$$ \frac{f(v)-f(u)}{v-u} \le \cfrac{f(w)-f(v)}{w-v} $$
This simply formalizes the intuition that adjacent chords on the graph of a convex function have non-decreasing slopes, and follows directly from the definition of convexity with $\lambda = \frac{w-v}{w-u} \in [0,1]$ and $1-\lambda=\frac{v-u}{w-u}\,$:
$$ \require{cancel} \begin{align} & f\big(\lambda u + (1-\lambda)w)\big) \le \lambda f(u) + (1-\lambda)f(w) \\ \iff \quad & f\left(\frac{(\cancel{w}-v)u+(v-\cancel{u})w}{w-u}\right) \le \frac{w-v}{w-u} f(u) + \frac{v-u}{w-u} f(w) \\ \iff \quad & f(v) \big((w-v)+(v-u)\big) \le (w-v)f(u) + (v-u)f(w) \\ \iff \quad & \big(f(v)-f(u)\big)(w-v) \le \big(f(w)-f(v)\big)(v-u) \end{align} $$
Proof. Since $f$ is increasing and not constant there exists $\exists \,c \in (a,b)$ such that $f(c) \gt f(a)$. For $\forall \,x_1,x_2$ such that $a\lt c\lt x_1\lt x_2\lt b$ applying the previous lemma twice gives:
$$ \frac{f(x_2)-f(x_1)}{x_2-x_1} \ge \frac{f(x_1)-f(c)}{x_1-c} \ge \frac{f(c)-f(a)}{c-a} \gt 0 $$
The above proves that $f$ is strictly increasing on $[c,b)$ since $x_1 \gt c \implies f(x_1) \gt f(c)$ and $x_2 \gt x_1 \gt c \implies f(x_2) \gt f(x_1)$.