Is it true that $\sup_{x \in \mathbb{R}} |f(x)| = \sup_{x \in \mathbb{Q}} |f(x)|$?

Question

Suppose $f \colon \mathbb{R} \rightarrow \mathbb{R}$ is a right-continuous function. Is it true that $\sup_{x \in \mathbb{R}} |f(x)| = \sup_{x \in \mathbb{Q}} |f(x)|$?

Answer 1

Suppose not. Then

$$\sup_{x \in \mathbb{R}} |f(x)|>\sup_{x \in \mathbb{Q}} |f(x)|$$

Now let $0<\varepsilon<\sup_{x \in \mathbb{R}} |f(x)|-\sup_{x \in \mathbb{Q}} |f(x)|$. This means there is a real (irrational) number $r$ such that for all $q\in\Bbb Q$, we must have $|f(r)|\ge|f(q)|+\varepsilon$. Now let

$$x_n=\tfrac{1}{n}\lfloor rn\rfloor$$

and see $r-x_n=\tfrac1n(nr-\lfloor rn\rfloor)$ so that (since $0\le nr-\lfloor rn\rfloor<1$) we know $0\le r-x_n<\tfrac1n$, and so clearly, $\lim x_n=r$, with every term $x_n$ being rational. Since each $x_n$ is rational, we have $|f(r)|\ge |f(x_n)|+\varepsilon$. Now we must have, since $f$ is continuous (and so is $|f|$):

\begin{align} |f(r)|&=\lim_{n\to\infty}|f(r)|\\ &\ge\lim_{n\to\infty}(|f(x_n)|+\varepsilon)\\ &=\varepsilon+\lim_{n\to\infty}|f(x_n)|\\ &=\varepsilon+|f(\lim_{n\to\infty}x_n)|\\ &=\varepsilon+|f(r)|\\ \end{align}

So now $|f(r)|\ge |f(r)|+\varepsilon$, which is a contradiction since $\epsilon>0$. Thus, the statement must be true:

$$\sup_{x \in \mathbb{R}} |f(x)|=\sup_{x \in \mathbb{Q}} |f(x)|$$

Answer 2

This is shorter:

Let $x\in\mathbb{R}$ and $(x_{n})\subseteq\mathbb{Q}$ converging to $x$ and such that $x_{n}\geq x$ (take $x_{n}\in[x,x+\frac{1}{n}]\cap\mathbb{Q}$, for example). Then, $|f(x)|=\lim|f(x_{n})|\leq \sup_{y\in\mathbb{Q}}|f(y)|$.