Proof: A strictly bandlimited signal must also be a signal of infinite time duration
A proof is given in the Wikipedia article on Bandlimiting:


  Assume that a signal \$f(t)\$ which has finite support in both domains exists. Let's sample it faster than the Nyquist frequency, and compute respective Fourier transform \$FT(f) = F_1(w)\$ and discrete-time fourier transform \$DTFT(f) = F_2(w)\$. According to properties of DTFT, \$F_2(w) = \sum_{n=-\infty}^{+\infty} F_1(w+n f_x)\$, where \$f_x\$ is the frequency used for discretization. If f is bandlimited, \$F_1\$> is zero outside of a certain interval, so with large enough \$f_x\$, \$F_2\$ will be zero in some intervals too, since individual supports of \$F_1\$ in sum of \$F_2\$ won't overlap. According to DTFT definition, \$F_2\$ is a sum of trigonometric functions, and since \$f(t)\$ is time-limited, this sum will be finite, so \$F_2\$ will be actually a trigonometric polynomial. All trigonometric polynomials are holomorphic on a whole complex plane, and there is a simple theorem in complex analysis that says that all zeros of non-constant holomorphic function are isolated. But this contradicts our earlier finding that \$F_2\$  has intervals full of zeros, because points in such intervals are not isolated. Thus the only time- and bandwidth-limited signal is a constant zero.