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This is a quick question directly related to the post How to show that $\sup\limits_{0<h<\delta}\frac{F(x+h)-F(x)}{h}$ is measurable for continuous $F$ on $\mathbb{R}$?.

There, for a continuous function $F:\mathbb{R}\to\mathbb{R}$ and $\delta>0$, one can make the reduction $$ \sup_{0<h<\delta}\frac{F(x+h)-F(x)}{h} = \sup_{h\in(0,\delta)\cap\mathbb{Q}}\frac{F(x+h)-F(x)}{h}. $$ for every $x$. To make this justification more clearly, can we say that the reason for this reduction is due to the fact that the function \begin{align*} G_x(h) := \frac{F(x+h) - F(x)}{h}, \end{align*} as a composition of continuous functions, is continuous on $(0, \delta)$ ?

shark
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2 Answers2

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Yes, I think that you’re right. In fact, as you already observed, the function $G_x$ is continuous in $(0, \delta)$, and since $(0, \delta) \cap \mathbb Q$ is a dense subset, the two suprema coincide by continuity.

To be more precise: let $X$ be a topological space, let $f \colon X \to \mathbb R$ be continuous and let $D \subset X$ be dense. Then, $\sup_D f = \sup_X f.$

Proof. Since $D \subset X$, we have that $\sup_D f \le \sup_X f$.

If $\sup_X f = +\infty$, for every $M > 0$ there exists $x \in X$ such that $f(x) > M$. Setting $\varepsilon := \frac{f(x)-M}{2} > 0$, it still holds that $f(x) -\varepsilon > M$. Since $f$ is continuous, there exists an open neighbourhood $U \ni x$ such that $\forall y \in U$, $| f(y) - f(x) | < \varepsilon$, so, in particular,$ f(y) - f(x) > -\varepsilon$, hence $f(y) > f(x) - \varepsilon > M$ $\forall y \in U$. Since $D$ is dense, $U \cap D \ne \emptyset$, so there exists $y_D \in D$ such that $f(y_D) > M$. Since this argument holds for every $M > 0$, we conclude that $\sup_D f = +\infty = \sup_X f$.

If $\sup_X f < +\infty$, fix an arbitrary $\eta > 0$. By the definition of supremum, there exists $x \in X$ such that $\sup_X f - \frac{\eta}{2} \le f(x) \le \sup_X f$. By continuity, there exists an open neighbourhood $U \ni x$ such that $|f(y) - f(x)| < \frac{\eta}{2}$ $\forall y \in U$. Arguing as above, by density one is able to find $y_D \in D \cap U$ such that $f(y_D) > f(x) -\frac{\eta}{2} \ge \sup_X -\eta$. Therefore, $\sup_D f \ge f(y_D) > \sup_X f -\eta$. Since $\eta > 0$ is arbitraty, we conclude that $\sup_D f \ge \sup_X f$, and this concludes the proof.

In your case, use this result with $X = (0, \delta)$ and $f = G_x$.

JaK
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Yepp, the reduction works because $G_x(h) = \frac{F(x+h)-F(x)}{h}$ is continuous on $(0,\delta)$ (as a quotient of continuous functions), and the supremum of a continuous function over an interval equals its supremum over any dense subset. Since rationals are dense in $(0,\delta)$, the suprema are equal.

Checking continuity of $G_x(h)$:

For $F$ continuous and $h_0 \in (0,\delta)$, I need $\lim_{h \to h_0} G_x(h) = G_x(h_0)$.

Since $G_x(h) = \frac{F(x+h) - F(x)}{h}$:

  • The numerator $F(x+h) - F(x)$ is continuous in $h$ (since $F$ is continuous).
  • The denominator $h$ is continuous and nonzero on $(0,\delta)$.
  • Therefore, their quotient is continuous on $(0,\delta)$.

Checking the supremum argument:

For a continuous function on an interval, the supremum equals the supremum over any dense subset.

Since $\mathbb{Q} \cap (0,\delta)$ is dense in $(0,\delta)$ and $G_x$ is continuous:

  • If $\sup_{h \in (0,\delta)} G_x(h) = M$, then for any $\epsilon > 0$, there exists $h_0$ with $G_x(h_0) > M - \epsilon$.
  • By continuity, $G_x(r) > M - \epsilon$ for rationals $r$ near $h_0$
  • Therefore $\sup_{h \in \mathbb{Q} \cap (0,\delta)} G_x(h) \geq M - \epsilon$ for all $\epsilon > 0$.
  • Hence the suprema are equal.

I guess the intuition is that continuity and density (between any two real numbers, you can always squeeze in a rational number) means that you can approximate the supremum arbitrarily well using just the dense subset.

arjo
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