Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a continuous function such that for all \(x \in \mathbb{R}\), \[ \int_0^1 f(xt) \, dt = 0. \quad \text{(*)} \] Then

• 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.

Hint:

When an integral condition holds for all \(x\), it often implies symmetry or cancellation properties in \(f\). In such cases, \(f\) can oscillate around zero instead of being identically zero.

Answer 1

The Correct Option is A

Step 1: Understanding the given condition.
We are told that for all \(x \in \mathbb{R}\), \[ \int_0^1 f(xt)\,dt = 0. \] Using the substitution \(u = xt\), we have \[ \frac{1}{x}\int_0^x f(u)\,du = 0 \quad \forall x \ne 0. \] Thus, \[ \int_0^x f(u)\,du = 0 \quad \forall x \in \mathbb{R}. \]
Step 2: Differentiating both sides.
Differentiating with respect to \(x\), we get \(f(x) = 0\) for all \(x\). However, we must also consider that differentiability of the integral condition is not assumed, only continuity of \(f\).
Step 3: Constructing a valid nontrivial function.
Consider an odd function \(f(x)\), e.g. \(f(x) = \sin(2\pi \log|x|)\) for \(x \ne 0\), and \(f(0)=0\). For such symmetric oscillatory functions, the integral from \(0\) to \(x\) can vanish for all \(x\). Hence \(f\) can take both positive and negative values and still satisfy the condition.
Step 4: Conclusion.
Thus, there exists an \(f\) that satisfies (*) and takes both positive and negative values.