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Sa determine functiile continue \( f:[0,\infty)\to\mathbb{R} \) cu proprietatea:

\( f(x)=\int_0^xf(x-t)f(t)dt,\forall x\in [0,\infty) \).

Let A>0 fixed. We work on [0,A]: let \( k=||f||_{[0,A]}=\sup_{[0,A]}|f| \).

\( |f*f(x)|\leq k^2x \)

By induction, the convolution of f n-times, with n>1 integer, gives

\( |f*f*...*f(x)|\leq \frac{k^{n}x^{n-1}}{(n-1)!} \)

\( f=f*f=f*f*...*f \)

Take supremum on [0,A] and obtain

\( k\leq \frac{k^{n}A^{n-1}}{(n-1)!} \)

Suppose k>0. Then \( 1\leq \frac{k^{n-1}A^{n-1}}{(n-1)!} \)

RHS tends to 0 when n tends to \( \infty \), therefore \( 1\leq 0 \), a contradiction.

We get k=0, so f =0 on [0,A] for A>0.