Prove that the right hand derivative can be found at any point in an open interval where $f$ is convex (without assuming differentiability)

Question

My book wanted a proof that the right hand derivative can be found at any point in an open interval where $f$ is convex (without assuming differentiability). This is the first time I used an infinimum as a limit point, so I just wanted to make sure I didn't break any rules.

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By the triple inequality lemma for a function $f$ that is (strictly) convex on an open interval $I$ (derived here: How to prove that convex function has an increasing slope?), we know that:

$$\forall x_1,x_2,x_3 \in I: x_1\lt x_2 \lt x_3 \rightarrow \frac{f(x_2)-f(x_1)}{x_2-x_1} \lt \frac{f(x_3)-f(x_1)}{x_3-x_1} \lt \frac{f(x_3)-f(x_2)}{x_3-x_2} \quad (*_1)$$

Now, choose some arbitrary $a \in I$ and hold it fixed. Then we have the following updated version of $(*_1)$:

$$\forall x_1,x_3 \in I: x_1\lt a \lt x_3 \rightarrow \frac{f(a)-f(x_1)}{a-x_1} \lt \frac{f(x_3)-f(x_1)}{x_3-x_1} \lt \frac{f(x_3)-f(a)}{x_3-a} \quad (*_2)$$

Because $I$ is an open interval, we can certainly find an $x_{\ell} \in I$ such that $x_{\ell} \lt a$. Then we can update $(*_2)$ to:

$$\forall x_3 \in I: x_{\ell} \lt a \lt x_3 \rightarrow \frac{f(a)-f(x_{\ell})}{a-x_{\ell}} \lt \frac{f(x_3)-f(x_{\ell})}{x_3-x_{\ell}} \lt \frac{f(x_3)-f(a)}{x_3-a} \quad (*_3)$$

Therefore, we can make the following claim:

$$\forall x_3\in I: x_3 \gt a \rightarrow \frac{f(a)-f(x_{\ell})}{a-x_{\ell}} \lt \frac{f(x_3)-f(a)}{x_3-a} \quad (*_4)$$

This means that the set defined by: $S=\left\{\frac{f(x_3)-f(a)}{x_3-a}: x_3\gt a\right\}$ has a lower bound. Because this argument takes place in $\mathbb R$, this means that the set has a greatest lower bound. Let $\beta=\text{inf}(S)$.

Our claim is that $\displaystyle \lim_{x_3 \to a^+}\frac{f(x_3)-f(a)}{x_3-a}=\beta$.

Before proving this, we require a useful lemma, which can be derived from the definition of convexity. For any $x_1,x_2,x_3 \in I$, we must have the following:

$$x_1 \lt x_2 \lt x_3 \rightarrow \frac{f(x_2)-f(x_1)}{x_2-x_1}\lt \frac{f(x_3)-f(x_1)}{x_3-x_1}\quad (\text{Definition})$$

Let $x_1 = a$. Then we have for any $x_2,x_3 \in I$, if $a \lt x_2 \lt x_3$

$$\frac{f(x_2)-f(a)}{x_2-a}\lt \frac{f(x_3)-f(a)}{x_3-a} \quad (*_5)$$

Now, consider an arbitrary $\varepsilon \gt 0$ and the value $\beta+\varepsilon$. Clearly, $\beta+\varepsilon \gt \beta$. Because $\beta=\text{inf}(S)$, we know that there must exist a $y \in S$ such that $\beta \lt y \lt \beta+\varepsilon$. This $y$ must be of the form $y=\frac{f(x_{\varepsilon})-f(a)}{x_{\varepsilon}-a}$. Further, because $y \in S$, we know that $x_{\varepsilon} \gt a$.

By $(*_5)$, we must have the following:

$$\forall x \in (a, x_{\varepsilon}): \frac{f(x)-f(a)}{x-a} \lt \frac{f(x_{\varepsilon})-f(a)}{x_{\varepsilon}-a}$$

This implies that for a $\delta = x_{\varepsilon}-a \gt 0$, we have that $\forall x \in (a,a+\delta): \frac{f(x)-f(a)}{x-a} \lt \varepsilon + \beta$, which implies that $\frac{f(x)-f(a)}{x-a}-\beta \lt \varepsilon$. Further, because $\beta$ is the infinimum of $S$, we must have that:

$$\forall x \in (a,a+\delta): \frac{f(x)-f(a)}{x-a}\gt \beta - \varepsilon \implies \frac{f(x)-f(a)}{x-a} - \beta \gt -\varepsilon$$

The previous two results show that for any $x \in (a,a+\delta) :\left|\frac{f(x)-f(a)}{x-a} - \beta\right| \lt \varepsilon$, which is precisely the definition of $\displaystyle \lim_{x \to a^+}\frac{f(x)-f(a)}{x-a}=\beta$

Answer 1

Your proof is correct. For more clarity, you can split it into two independent parts:

Part 1: Let $I \subset \Bbb R$ be an open interval and $f: I \to \Bbb R$ a convex function. Then $$ \forall a, x, y, z \in I, x < a < y < z \implies \frac{f(x)-f(a)}{x-a} \le \frac{f(y)-f(a)}{y-a} \le \frac{f(z)-f(a)}{z-a} \, . $$

This shows that the function $$ I \cap (a, \infty) \to \Bbb R, x \mapsto \frac{f(x)-f(a)}{x-a} $$ is increasing and bounded below.

Part 2: Let $g: (a, b) \to \Bbb R$ be a function which is increasing and bounded below. Then $$ \lim_{x \to a^+} g(x) = \inf \{ g(x) \mid x \in (a, b) \} \, . $$ In particular, the limit on the left exists.

The proof of the second part can be simplified at bit: Let $ \beta = \inf \{ g(x) \mid x \in (a, b) \}$ and $\epsilon > 0$. Then there is a $y \in (a, b)$ such that $g(y) < \beta + \epsilon$. It follows that for all $x \in (a, y)$ $$ \beta \le g(x) \le g(y) < \beta + \epsilon \implies |g(x) - \beta| = g(x) - \beta < \epsilon \, . $$