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Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that for all $x \in \mathbb{R}$,

$\int_{0}^{1} f(xt) dt = 0. \quad (*)$

Then which of the following options is correct?

(A) $f$ must be identically 0 on the whole of $\mathbb{R}$.

(B) There is an $f$ satisfying (*) that is identically 0 on $(0, 1)$ but not identically 0 on the whole of $\mathbb{R}$.

(C) There is an $f$ satisfying (*) that takes both positive and negative values.

(D) There is an $f$ satisfying (*) that is 0 at infinitely many points but is not identically zero.
in MA2021 | 164 views

(A) $f$ must be identically 0 on the whole of $\mathbb{R}$.

Explanation:

Let's proceed with proof by contradiction.

Assume that there exists a continuous function $f$ satisfying (*), but $f$ is not identically zero on $\mathbb{R}$.

Since $f$ is not identically zero, there exists $c \in \mathbb{R}$ such that $f(c) \neq 0$. Without loss of generality, let's assume $f(c) > 0$.

Since $f$ is continuous, there exists an interval $(a, b)$ containing $c$ such that $f(x) > 0$ for all $x \in (a, b)$.

Now, consider the integral:

$\int_{0}^{1} f(tx) dt$

Let $u = tx$, then $du = x dt$, and the integral becomes:

$\int_{0}^{x} \frac{1}{x} f(u) du$

Using the given condition (*), this integral is equal to 0:

$\int_{0}^{x} \frac{1}{x} f(u) du = 0$

Since $f(u) > 0$ for all $u \in (a, b)$, and the integrand is positive, the integral must be positive for all $x \in (0, 1)$:

$\int_{0}^{x} \frac{1}{x} f(u) du > 0$ for all $x \in (0, 1)$

But this contradicts the fact that the integral is equal to 0 for all $x \in (0, 1)$ according to the condition (*).

Hence, our assumption that there exists a continuous function $f$ satisfying (*) but not identically zero on $\mathbb{R}$ is incorrect.

Therefore, the only possibility is that $f$ must be identically zero on the whole of $\mathbb{R}$.

Hence, option (A) is correct: $f$ must be identically 0 on the whole of $\mathbb{R}$.

Note: This solution is AI-generated but verified by an expert.

by Professor