This is about the Fourier uncertainty principle, which is closely related to, but not the same as the Heisenberg uncertainty principle in physics. So no $\hbar$.
This may be expressed with scaling and definitions more preferred by electrical engineers than might be the habit of mathematicians.
The definition of the continuous Fourier Transform preferred by most electrical engineers:
$$ X(f) = \mathscr{F} \Big\{ x(t) \Big\} = \int\limits_{-\infty}^{+\infty} x(t) \, e^{-i 2 \pi f t} \ \mathrm{d}t $$
and inverse:
$$ x(t) = \mathscr{F}^{-1} \Big\{ X(f) \Big\} = \int\limits_{-\infty}^{+\infty} X(f) \, e^{+i 2 \pi f t} \ \mathrm{d}f $$
Even with different signs on $i$, the elegant symmetry between the forward transform and inverse should be clear. And it makes remembering the Duality property:
If $X(f) = \mathscr{F} \Big\{ x(t) \Big\}$, then $$x(-f) = \mathscr{F} \Big\{ X(t) \Big\}$$
the "DC values":
$$ X(0) = \int\limits_{-\infty}^{+\infty} x(t) \ \mathrm{d}t $$
$$ x(0) = \int\limits_{-\infty}^{+\infty} X(f) \ \mathrm{d}f $$
and Parseval's theorem:
$$ E = \int\limits_{-\infty}^{+\infty} \Big| x(t) \Big|^2 \ \mathrm{d}t = \int\limits_{-\infty}^{+\infty} \Big| X(f) \Big|^2 \ \mathrm{d}f$$
…easy. $E$ is the total energy of the signal in both the time or frequency domains.
No nasty asymmetrical scaling factors to worry about! (Just remember the $2\pi$ in the exponent.) This is why EE's like this definition of the Fourier Transform.
If you define the the square of the time bandwidth of $x(t)$ as
$$ w^2 = \frac{1}{E} \int\limits_{-\infty}^{+\infty} t^2 \Big| x(t) \Big|^2 \ \mathrm{d}t $$
and, using the above definition, the square of the frequency bandwidth of $X(f)$ as
$$ W^2 = \frac{1}{E} \int\limits_{-\infty}^{+\infty} f^2 \Big| X(f) \Big|^2 \ \mathrm{d}f $$
and toss in this condition in the limit,
$$ \lim_{|t|\to\infty} \sqrt{|t|} \ x(t) = 0 $$
Then,
$$ w \cdot W \ge \frac{1}{4 \pi} $$
In consistent units (like time in seconds and frequency in hertz), then the time bandwidth times the frequency bandwidth must equal or exceed $\frac{1}{4 \pi}$.
I could derive this here if someone requires (it requires Cauchy-Schwarz inequality and some messing around), but my question is about something that puzzles me. Suppose I delay $x(t)$ by some large, but finite delay, $\tau$
$$ x_\tau(t) = x(t-\tau) $$
That certainly that increases the time bandwidth measure
$$\begin{align}
w_\tau^2 &= \frac{1}{E} \int\limits_{-\infty}^{+\infty} t^2 \Big| x_\tau(t) \Big|^2 \ \mathrm{d}t \\
&= \frac{1}{E} \int\limits_{-\infty}^{+\infty} t^2 \Big| x(t-\tau) \Big|^2 \ \mathrm{d}t \\
&= \frac{1}{E} \int\limits_{-\infty}^{+\infty} (t+\tau)^2 \Big| x(t) \Big|^2 \ \mathrm{d}t \\
\end{align} $$
There is no assumption that the mean $t$ is zero. Perhaps $ \int_{-\infty}^{+\infty} t \Big| x(t) \Big|^2 \ \mathrm{d}t \ne 0 $.
This is like measuring the mean-square $t$, not the variance of $t$.
But delaying by $\tau$ does nothing to the magnitude of the Fourier Transform:
$$ X_\tau(f) = \mathscr{F} \Big\{ x_\tau(t) \Big\} = \mathscr{F} \Big\{ x(t-\tau) \Big\} $$
…
$$ \big| X_\tau(f) \big| = \big| X(f) \big| $$
So the frequency bandwidth $W_\tau$ is not affected:
$$\begin{align}
W_\tau^2 &= \frac{1}{E} \int\limits_{-\infty}^{+\infty} f^2 \Big| X_\tau(f) \Big|^2 \ \mathrm{d}f \\
&= \frac{1}{E} \int\limits_{-\infty}^{+\infty} f^2 \Big| X(f) \Big|^2 \ \mathrm{d}f \\
&= W^2 \\
\end{align} $$
It doesn't break the inequality, but wouldn't I expect $W_\tau<W$ if $w_\tau>w$? Do we also need to require
$$ \int\limits_{-\infty}^{+\infty} t \Big| x(t) \Big|^2 \ \mathrm{d}t = 0 $$
and
$$ \int\limits_{-\infty}^{+\infty} f \Big| X(f) \Big|^2 \ \mathrm{d}f = 0 $$
so that both $x(t)$ and $X(f)$ are centered about $t=0$ and $f=0$ and the time width and frequency width cannot be artificially increased just by translating in time or frequency?
Best Answer
Let’s say that $w$ is as small as possible when $\tau$ is 0. Then $w \cdot W$ satisfies the inequality. Delaying it, as you have by letting $\tau\neq 0$, increases $w$ and therefore increases $w \cdot W$, so the inequality is still valid.
An insight from your thoughts is that $w \cdot W$ satisfies the inequality even if you’re allowed to shift the function in time so that $w$ is as small as possible.