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Prove if $f : (a,b) \to \Bbb R$ is a twice-differentiable, strictly decreasing convex function then $f$ is bounded from below.

I tried using the mean value theorem but I couldn't get any information about the behavior of $f$.

Kantig Shoter
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3 Answers3

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It is no loss of generality to assume that $f:[a',b)\to \mathbb R$ for some $a<a'<b$ and in fact that $a'=0$ and $f(a')=0.$ Then, for any $0<b'<b,\ f\ $ has a finite minimum on $[0,b']$ so if $f$ is not bounded below, then there exists a sequence $(x_n)$ increasing to $b$ such that $f(x_n)\to -\infty.$

Convexity of $f$ implies that for $x<y<z,$ if $\varphi(w) = \frac{f(w) - f(x)}{w - x}$ then $\varphi(y)\le \varphi(z).$ Applying this to our situation, we have $x=b'=f(b')=0$ so $\varphi(w)=\frac{f(w)}{w}$ is increasing on $[0,b).$ But then $\varphi(x_n)$ is an increasing sequence with $\varphi(x_n)<0$ so it is bounded, at least eventually: there is an integer $N$ such that $\varphi(x_N)\neq 0.$ Then $\varphi(x_N)\le \varphi(x_n)\le 0$ as soon as $n\ge N.$ This is a contradiction because $f(x_n)\to -\infty$ and $x_n\to b.$

Matematleta
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  • I am not extending $f$, rather restricting its domain from $(a,b)$ to a smaller subset of $(a,b).$ If $f$ is strictly decreasing on $(a,b)$ then if we choose an $a<a'<b$ if $f$ is bounded below on $[a', b)$ then it is also bounded below on $(a,b).$ – Matematleta Dec 28 '22 at 14:24
  • I don't see how you get a contradiction. You have $f(x_{n+1}) \ge {x_{n+1} \over x_n} f(x_n)$, but since $f(x_n) <0$, you cannot immediately conclude that $f(x_{n+1}) > f(x_n)$. – copper.hat Dec 28 '22 at 19:00
  • Convexity of $f$ implies that for $x<y<z,$ if $\varphi(w) = \frac{f(w) - f(x)}{w - x}$ then $\varphi(y)\le \varphi(z).$ Applying this to our situation, we have $x=b'=f(b')=0$ so $\varphi(w)=\frac{f(w)}{w}$ is increasing on $[0,b).$ But then $\varphi(x_n)$ is an increasing sequence with $\varphi(x_n)<0$ so it is bounded, at least eventually: there is an integer $N$ such that $\varphi(x_N)\neq 0.$ Then $\varphi(x_N)\le \varphi(x_n)\le 0$ as soon as $n\ge N.$ This is a contradiction because $f(x_n)\to -\infty$ and $x_n\to b.$ – Matematleta Dec 29 '22 at 16:54
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Let $c={1 \over 2} (a+b)$ below.

If $f$ is any real valued convex function on $(a,b)$ then it is bounded below. Pick any subgradient $g$ at $c$, then $f(x)-f(c) \ge g(x-c) \ge -|g|{1 \over 2} (b-a)$.

Alternative that does not involve subgradients:

An important property of convex functions on the real line is that the secant function $R$ in https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable is symmetric and non decreasing as a function $x \mapsto R(x,x_2)$.

A little bit of work shows that $g_+ = \lim_{h \downarrow 0} R(c,c+h)$ exists and since $g_+ \le R(c,x) =R(x,c)$, we have $f(x) - f(x) \ge g_+(x-c)$ for $c<x$.

Repeating shows that $g_- = \lim_{h \uparrow 0} R(c,c+h)$ exists (note that $h$ is negative) and as above we have $R(c,x) = R(x,c) \le g_-$ for $x<c$ and this gives $f(x) - f(c) \ge g_-(x-c)$ for $x<c$.

The properties of $R$ show that $g_- \le g_+$. (As an aside, the elements of $[g_-,g_+]$ are the subgradients of $f$ at $c$.)

Combining these (and a little computation) shows that $f(x) -f(c) \ge - \max(|g_-|,|g_+|){1 \over 2} (b-a)$ and hence $f$ is bounded below.

copper.hat
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Well, this is more of an intuition than an answer. If $f(x)$ is a convex function in $(a,b)$, so it implies its first derivative is increasing, so $f''(x) \ge 0, x \in (a,b)$. But, we are also given that the function is strictly decreasing, meaning $f'(x)\le0, x \in (a,b)$. So, with a bit of thought (again note this is not a complete proof), for the function to remain convex yet decreasing in $(a,b)$ we can show that $$f''(x) \to 0$$ and $$f'(x) \to 0$$ so it is bounded from below. This is because the rate of change of the first derivative must decrease (yet not become negative otherwise function is concave) otherwise the first derivative will become positive (so the function becomes increasing).

D S
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