Multiplier for the space of functions which have a primitive

Solution 1:

Let us prove that if $g$ has the "multiplier property", then it is essentially bounded. Clearly, $g^2$ also has the "multiplier property", and $g$ is bounded if and only if $g^2$ is bounded, so with no loss of generality we may assume that $g \ge 0$. Suppose, contrary to our claim, that $g$ is not essentially bounded. We will prove that there is a function $f$ such that $f \in Pr([a,b])$, but $f g \notin Pr([a,b])$.

Step 1. Almost every point of $[a, b]$ is a Lebesgue point of $g$. Hence, for each $n = 1, 2, \ldots$ there is a Lebesgue point $x_n$ of $g$ such that $g(x_n) \ge n$. By passing to a subsequence if necessary, we may assume that $x_n$ is monotone – and thus it converges to some limit $x_0$ – and that $|x_{n+1} - x_0| < \tfrac{1}{n+1} |x_n - x_0|$. With no loss of generality we may assume that $[a, b]$ contains $0$, $x_n$ is decreasing and $x_0 = 0$; thus, $x_{n+1} < \tfrac{1}{n+1} x_n$.

Step 2. Choose $\varepsilon_n > 0$ so that $\varepsilon_n < \tfrac{1}{4} x_n$, and furthermore $$ \tag{1} \operatorname{Leb} \{x \in (x_n + \varepsilon_n, x_n + 3 \varepsilon_n) : g(x) \ge \tfrac{1}{2} g(x_n) \} > \varepsilon_n $$ (this is possible, because $x_n$ is a Lebesgue point of $g$). Next, choose $\alpha_n > 0$ in such a way that $$ \tag{2} \frac{1}{x_k} \sum_{n = k}^\infty \alpha_n \varepsilon_n g(x_n) = 4 $$ (that is, $\alpha_n = 4 (x_n - x_{n+1}) / (\varepsilon_n g(x_n))$). Then clearly $$ \tag{3} \frac{1}{x_k} \sum_{n = k}^\infty \alpha_n \varepsilon_n \le \frac{4}{\inf \{g(x_n) : n \ge k\}} \le \frac{4}{k} \, . $$

Step 3. Define $f$ to be a function which is nonnegative, equal to $0$ outside the union of $(x_n, x_n + 4 \varepsilon_n)$, smooth except at $x_0$, and such that $$ 0 \le f(x) \le \alpha_n $$ when $x \in (x_n, x_n + 4 \varepsilon_n)$, and $$ f(x) = \alpha_n $$ when $x \in (x_n + \varepsilon_n, x_n + 3 \varepsilon_n)$.

Step 4. Let $F$ be the indefinite Lebesgue integral of $f$. Clearly, $F'(x) = f(x)$ for all $x$ except perhaps $x = 0$. However $F'_-(0) = 0$, and if $$ x_k \le x \le x_{k-1} ,$$ then, by (3), $$ 0 \le \frac{F(x) - F(0)}{x - 0} \le \frac{1}{x_k} \sum_{n = k}^\infty \int_{x_n}^{x_n + 4 \varepsilon_n} f(x) dx \le \frac{1}{x_k} \sum_{n = k}^\infty 4 \alpha_n \varepsilon_n \le \frac{16}{k} \, .$$ Therefore, $F'_+(0) = 0$, and hence $F'(0) = 0 = f(0)$. Therefore, $f \in Pr([a, b])$.

Step 5. On the other hand, let $H$ be the indefinite Lebesgue integral of $f g$. We claim that $H$ is not differentiable at $x = 0$. By (1) we have $$ \int_{x_n + \varepsilon_n}^{x_n + 3 \varepsilon_n} f(x) g(x) dx \ge \frac{\alpha_n \varepsilon_n g(x_n)}{2} \, , $$ so that, by (2), $$ \frac{H(x_k + 4 \varepsilon_k) - H(0)}{(x_k + 4 \varepsilon_k) - 0} \ge \frac{1}{2 x_k} \sum_{n = k}^\infty \frac{\alpha_n \varepsilon_n g(x_n)}{2} = 1. $$ Therefore, $$ \limsup_{x \to 0^+} \frac{H(x) - H(0)}{x - 0} \ge 1 . $$ If $a < 0$, then $F'_-(0) = 0$, and we are done.

Step 6. The case $a = 0$ requires an additional calculation. In addition to (1) we need to assume that $$ \int_{x_n}^{x_n + 4 \varepsilon_n} g(x) dx \le 8 \varepsilon_n g(x_n) $$ (again this is possible because $x_n$ is a Lebesgue point of $g$). Then $$ \int_{x_n}^{x_n + 4 \varepsilon_n} f(x) g(x) dx \le 8 \alpha_n \varepsilon_n g(x_n) \, , $$ so that by (2), $$ \frac{H(x_{k-1}) - H(0)}{x_{k-1} - 0} \le \frac{1}{x_{k-1}} \sum_{n = k}^\infty 8 \alpha_n \varepsilon_n g(x_n) = \frac{32 x_k}{x_{k-1}} \le \frac{32}{k} \, . $$ Therefore, $$ \liminf_{x \to 0^+} \frac{H(x) - H(0)}{x - 0} \le 0 . $$ It follows that $F'_+(0)$ does not exist.