This is (the first half of) exercise 14 in Baby Rudin
Let $f:(a, b) \to \mathbb{R}^1$ be differentiable. Prove that $f$ is convex iff $f'$ is monotonically increasing.
($\Rightarrow$) Assume $f$ is convex in $(a, b)$. Fix $0 < \lambda < 1$ and $a < y \le x < b$. Notice that
\begin{align}\tag{13.1}
y \le x \implies y (1-\lambda) \le x (1-\lambda) \implies y – \lambda y \le x -\lambda x \implies \lambda x + y – \lambda y \le x
\end{align}
By the definition of convexity, we have that
\begin{equation}\tag{13.2}
f[\lambda x + y – \lambda y] \le \lambda f(x) + f(y) – \lambda f(y)
\end{equation}
and differentiating (13.2) with respect to x, we have
\begin{equation}\tag{13.3}
f^{\prime}[\lambda x + y – \lambda y] \cdot \lambda \le \cdot \lambda f^{\prime}(x) \implies f^{\prime}[\lambda x + y – \lambda y] \le f^{\prime}(x)
\end{equation}
By (13.1) and (13.3), we conclude that $f'$ is monotonically increasing.
($\Leftarrow$) Suppose $f'$ is monotonically increasing. Fix $0< \lambda< 1$ and suppose $f$ is not convex. Then $\exists p, q \in (a, b)$ such that
\begin{equation}\tag{13.4}
f(\lambda p + q – \lambda q) > \lambda f(p) +f(q) – \lambda f(q) \stackrel{\textrm{w.r.t. } p}{\implies} f'(\lambda p + q- \lambda q) > f'(p)
\end{equation}
Without loss of generality, let $p \geq q$, which implies
\begin{align*}
p (1-\lambda) > q (1-\lambda) \implies p – \lambda p > q -\lambda q \implies p > \lambda p + q – \lambda q
\end{align*}
Since $f'$ is monotonically increasing, we get $f'(p) > f'(\lambda p + q- \lambda q)$ which contradicts (13.4).
Can someone please critique my proof? Please don't bother suggesting a new proof as those can be found here and here. I am new to handling derivatives in an abstract setting, so I am not sure if it is valid to differentiate (13.3) and preserve the direction of the inequality, like I did. Is there a theorem/ lemma that supports this move?
Best Answer
You can usually never differentiate inequalities.
If $f,g : ]a,b[ \rightarrow \mathbb R$ are differentiable, $$ \forall x \in ]a,b[, \ f(x) \leq g(x) $$ does not imply $$ \forall x \in ]a,b[, \ f'(x) \leq g'(x). $$
For example, take $f(x) = x^2$ and $g(x) = x$. We have $f(x) \leq g(x)$ on $[\frac{1}{2},1]$ and yet for all $x \in [\frac{1}{2},1]$, $f'(x) = 2x \geq 1 = g'(x)$.
The reason the result fails is because saying a function is above another does not tell us anything about how fast these functions grow comparatively.
You can try to draw a counter-example to convince yourself.
Hope this helps!