I am following up on my previous question. My previous attempt for the proof was wildly incorrect (my question was how that proof was exactly my old proof was incorrect) and I have now come up with a new proof.
I have to prove:
Let $f:(a, b) \to R^1$ be differentiable. Prove that $f$ is convex iff $f'$ is monotonically increasing.
What I have for the proof:
($\Rightarrow$) Assume $f$ is convex in $(a, b)$. Let $a<s<t<u<b$. By Exercise 23 in Chapter 4,
\begin{align}\tag{14.1}
\frac{f(t)-f(s)}{t-s} \le \frac{f(u)-f(s)}{u-s} \le \frac{f(u)-f(t)}{u-t}
\end{align}
Since $f$ is differentiable on $(a,b)$, both $ f'(s) = \lim_{t \to s} \frac{f(t)-f(s)}{t-s}$ and $f'(t)=\lim_{u \to t} \frac{f(u)-f(t)}{u-t}$ exist. However, applying the Order Limit Theorem on (14.1) gives
\begin{align*}
\lim_{t \to s} \frac{f(t)-f(s)}{t-s} \le \lim_{u \to t} \frac{f(u)-f(t)}{u-t} \implies f'(s) \le f'(t)
\end{align*}
which shows that $f'$ is monotonically increasing in $(a, b)$.($\Leftarrow$) Assume $f'$ is monotonically increasing in $(a, b)$ and $a<x<y<b$. Fix $0 < \lambda< 1$. By Exercise 23 in Chapter 4, we must show that
\begin{equation}\tag{14.0}
f(\lambda x + (1- \lambda)y) \le \lambda f(x) + (1-\lambda) f(y)
\end{equation}
Denote $z=\lambda x+ (1-\lambda)y$.Then, $z=\lambda(x-y)+y$ which implies that $\lambda=\frac{z-y}{x-y}$. Since $\lambda>0, z-y>x-y \implies z>x$. Also, $1-\lambda=\frac{x-y-z+y}{x-y} = \frac{x-z}{x-y}$. Since $\lambda<1, x-z>x-y \implies z < y$. Thus, $x<z<y$. Then, (14.0) can be simplified as:
\begin{align*}
f(z) &\le f(y) + \lambda f(x) – \lambda f(y) \\
\lambda f(z) – \lambda f(x) &\le f(y) – f(z) – \lambda f(y) + \lambda f(z) \\
\lambda[f(z)-f(x)] &\le (1-\lambda)[f(y)-f(z)]
\end{align*}
Thus, since $\lambda = \frac{y-z}{y-x}$ and $1-\lambda = \frac{z-x}{y-x}$, it suffices to show that
\begin{equation}\tag{14.2}
\frac{f(z)-f(x)}{z-x} \le \frac{f(y)-f(z)}{y-z}
\end{equation}
Now, as we take $z\to x$ on the left of (14.2) and $y\to z$ on the right of (14.2), then we have $f'(x)\le f'(z)$, which holds since $x<z$ and $f'$ is monotonically increasing.
Exercise 23 in Chapter 4 in Rudin:
A real-valued function $f$ defined in $(a, b)$ is said to be convex if
$$ f \left( \lambda x + (1- \lambda) y \right) \leq \lambda f(x) + (1-\lambda) f(y)$$
whenever $a < x < b$, $a < y < b$, $0 < \lambda < 1$. Prove that every convex function is continuous.Hint: If $f$ is convex in $(a, b)$ and if $a < s < t < u < b$, show that
$$ \frac{ f(t)-f(s)}{t-s} \leq \frac{ f(u)-f(s)}{u-s} \leq \frac{ f(u)-f(t)}{u-t}.$$
Can someone please read over my proof and see if there is something that I did incorrectly? Also, specifically, is my usage of the Order Limit Theorem correct and is the argument right below (14.2) correct?
Best Answer
Hint: (reverse implication)
If $s<t<u$, then by the mean value theorem there exists $\xi_1 \in (s,t)$ and $\xi_2 \in (t,u)$ such that (since $f'$ is monotonically increasing) $$\frac{f(t)-f(s)}{t-s} = f'(\xi_1) \leqslant f'(\xi_2) = \frac{f(u)-f(t)}{u-t}$$
Forward Implication
By convexity, for $s < t < u$, we have
$$\frac{f(t)-f(s)}{t-s} \leqslant \frac{f(u)-f(s)}{u-s} \leqslant \frac{f(u)-f(t)}{u-t}$$
Thus,
$$f'(s) = \lim_{t \to s+}\frac{f(t)-f(s)}{t-s} \leqslant\lim_{t \to s+}\frac{f(u)-f(t)}{u-t} = \frac{f(u)-f(s)}{u-s}, $$
and
$$\frac{f(u)-f(s)}{u-s} = \lim_{t \to u-}\frac{f(t)-f(s)}{t-s} \leqslant \lim_{t \to u-} \frac{f(u)-f(t)}{u-t} = f'(u)$$
Therfore, $f'(u) \geqslant f'(s)$ when $u > s$ and $f$ is monotonically increasing.