Function satisfies its own integral equation

Calculus Level 3

Find all differentiable functions that satisfy $f(x) = \int_0^1 f(tx) dt$

It is a trivial exercise to show that the set of these functions forms a subspace of the vector space of all differentiable functions. Select as your answer the dimension of this vector space.

Extra: What happens if instead of integrating over $[0,1]$ we consider other intervals like $[ a,b ]$ or $( -\infty, \infty )$ ?

0 1 2 3

\infty

1 solution

Leonel Castillo
Nov 5, 2018

Given that $f$ is differentiable, take its derivative: $f'(x) = \int_0^1 t f'(tx) dt$ The left hand side is in perfect condition to apply integration by parts. Take $u = t, dv = f'(tx) dt$ . We obtain $f'(x) = \frac{f(x)}{x} - \int_0^1 \frac{f(tx)}{x} dt$ But notice that because of the given integral equation, $\int_0^1 \frac{f(tx)}{x} dt = \frac{\int_0^1 f(tx) dt}{x} = \frac{f(x)}{x}$ . Thus $f'(x) = 0$ . Integrating we obtain the general solution $f(x) = C$

So the solutions are the set of all constant functions, which in the given vector space has dimension 1.

Function satisfies its own integral equation

1 solution

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