Sa determine functiile continue \( f:[0,\infty)\to\mathbb{R} \) cu proprietatea:
\( f(x)=\int_0^xf(x-t)f(t)dt,\forall x\in [0,\infty) \).
Integrala "convolutata"
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Integrala "convolutata"
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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.
\( |f*f(x)|\leq k^2x \)
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.