In the notes I am reading they use the following. Let $U$ be the voltage (depends on time) and $I$ the current in a circuit with some resistor with resistance $R$. Then the power is given by:
$$P(t)=U(t)I(t).$$
This is then Fourier transformed to the following:
$$P(\omega)=U(\omega)I^*(\omega),$$
where $*$ denotes complex conjugated. Could somebody give a proof of this?
I think you are confusing complex power with instantaneous power.
The instantaneous power is simply $p(t)=v(t)i(t)$ as you've mentioned. The complex power however is not the Fourier transform of $P$. For two harmonic signals with frequency $\omega$, The complex power is instead defined as:
$$S:=\frac 12 VI^*$$
where $V$ and $I$ are the phasors of the voltage and current, respectively. Written in rms values $V_\mathrm{rms}=V/\sqrt2$ and $I_{rms}=I/\sqrt2$, this is simply:
$$S = V_\mathrm{rms}I^{*}_{\mathrm{rms}}$$
which is what you're referring to as $P=UI^*$.
But if these quantities are not the same, why are they both called power? This is simply because if I take the real part of the complex power, I get the time -averaged instantaneous power. To see why, let $v(t) = |V| \cos(\omega t + \phi_v)$ and $i(t) = |I| \cos(\omega t + \phi_i)$ be our two harmonic voltage and current signals. the time-averaged power is:
$$P_\mathrm{avg} = \frac{1}{T}\int_{0}^T v(t)i(t)dt \\
= \frac{1}{T}\int_{0}^T |V| \cos(\omega t + \phi_v)\times |I| \cos(\omega t + \phi_i)dt \\
= \frac 12 |V||I| \cos(\phi_v-\phi_i) \\
=\frac 12 \mathrm{Re}(|V| e^{-i\phi_v}|I|e^{-i\phi_i}) \\
= \frac 12 \mathrm{Re}(VI^*)$$
which finally gives:
$$P_\mathrm{avg} = \mathrm{Re}(S)$$
Note that the imaginary part of $S$ is usually called the reactive power, which shows a constant back-and-forth transmission of power between the source and the load.
In conclusion, the complex power $S$ is not the Fourier transform of the instantaneous power. As hyportnex and David point out in the comments, Taking the Fourier transform of $v(t)i(t)$ gives $V(\omega)*I(\omega)$, where $*$ is understood as convolution.
You may be curious about what happens if you use a Fourier transform instead of phasors in the above definition of $S$. Even though these two concepts are related, they are not the same. Even dimensionally, the phasor $V$ of a voltage has units of Volts, while its Fourier transform has units of Volts/Hz.
You can easily see that the quantity $\varepsilon(\omega) = V(\omega)I^*(\omega)$ is nothing but the energy spectral density. where $V(\omega)$ and $I(\omega)$ are the Fourier transforms of $v$ and $i$, respectively (as opposed to phasors).
This is an immediate consequence of Parseval's theorem, which says: $$\int_{-\infty}^{\infty}\varepsilon(\omega) d\omega = \int_{-\infty}^{\infty} V(\omega)I^*(\omega) d\omega \\ = \int_{-\infty}^{\infty} v(t)i(t) dt = \int_{-\infty}^{\infty} p(t) dt= E$$ which is the total energy.
