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In signal processing, there is a time-frequency duality. Band-limited signals have infinite time duration and time-limited signals have infinite bandwidth.
Where can I find a proof of this and is there some intuitive explanation for this?
I am not sure if it has to do something with fouriers uncertainty principle?
There are several proofs of this important property, e.g., on this wikipedia page on band-limiting, and on this dsp blog page. I haven't checked them in detail, so it's up to you to decide whether these proofs are sufficiently rigorous.
I want to show another proof which is very simple, if you accept the truth of the Paley-Wiener condition:
A square-integrable function $A(\omega)\ge 0$ is the magnitude of the Fourier transform of a causal function if and only if $$\int_{-\infty}^{\infty}\frac{|\log A(\omega)|}{1+\omega^2}d\omega<\infty\tag{1}$$
Assume that $f(t)$ is a time-limited function, i.e., $f(t)=0$ for $|t|>T$, and $F(\omega)$ is its Fourier transform. Then $f(t-T)$ is causal and its Fourier transform is $F(\omega)e^{-j\omega T}$. According to the Paley-Wiener condition for causal functions, $A(\omega)=|F(\omega)e^{-j\omega T}|=|F(\omega)|$ cannot be zero for $|\omega|>\Omega$, regardless of the choice of $\Omega$, because otherwise the integral in $(1)$ is not finite. Consequently, $f(t)$ cannot be band-limited.
Note that this proof shows that a time-limited function cannot be band-limited, but it also shows that a band-limited function cannot be time-limited, because if the inverse Fourier transform of a band-limited function were time-limited, it could be made causal by an appropriate shift in the time domain, which would not change its magnitude spectrum. So it would need to satisfy the Paley-Wiener condition, which we've shown to be impossible for a band-limited function.
Here is an argument that doesn't require the uncertainty principle.
Assume $y(t)$ is a signal that is not identical to zero, with Fourier transform $Y(f)$. Let $s(t) = \text{rect}(t/T)$ a rectangular pulse of duration $T$ and Fourier transform $S(f) = |T| \text{sinc}(T f)$.
Then, $x(t) = y(t)s(t)$ is a time-limited signal with Fourier transform $X(f) = Y(f) \ast S(f)$. Since $S(f)$ has infinite support, so does $X(f)$.
Since any time-limited signal $x(t)$ can be defined in this way for a suitable choice of $y(t)$ and $s(t)$ (and a possible time-shift that doesn't affect the result), then we can say that all such signals have infinite bandwidth.
The same argument can be applied for the inverse Fourier transform: band-limiting a signal involves multiplication by a rectangular pulse in the frequency domain, or a convolution with a sinc in the time domain, resulting in an infinite-duration signal.
