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Using the Taylor series expansion we have (for a sufficiently regular function $f$): $$ f(x+a)=\sum_{k=0}^n \frac{f^{(k)}(x)a^k}{k!} $$ So, defining the differential operator $D=\frac{d}{dx}$ and using the series expansion definition of the exponential function, we can write: $$ S_a f(x)=\exp(aD) f(x) $$ where $S_a f(x)=f(x+a)$ is the shift operator.

This gives an ''intuitive'' meaning to the exponential of the differential operator : $\exp(aD)=S_a$.

We can extend this intuition without problems and say that: $\log S_a=aD$ ?

And we can also found one (or more) operator $T=\log (aD)$, such that $\exp T=aD$?

I have found many references to the similar question for linear operators in finite dimensional vector space, but very little about the infinite dimensional case. Someone know some good reference in the web?

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At least from one perspective, what you're looking at is something called functional calculus, which is the business of defining and computing with functions of operators on vector spaces, typically (infinite-dimensional) Hilbert spaces. However, if you're specifically interested in functions of operators constructed from (partial) differentiation, then there is a specialised theory called the theory of pseudo-differential operators, which, very roughly speaking, uses the Fourier transform to "diagonalise" (partial) differentiation as an operator on a suitable function space.

So, let $\hat{f}$ denote the Fourier transform of a (sufficiently well-behaved) function $f$ and let $\check{g}$ denote the inverse Fourier transform of a (sufficiently well-behaved) function $g$, so that $$ \hat{f}(k) = \int_{-\infty}^\infty f(x)e^{-2\pi i kx}\,dx, \quad \check{g}(x) = \int_{-\infty}^\infty g(k) e^{2\pi i kx}\,dk. $$ If $f$ is a rapidly-decreasing smooth function on the real line, then, it's a basic property of the Fourier transform that $$ \widehat{Df}(k) = 2\pi i k \hat{f}(k), $$ or equivalently, that $$ Df(x) = \int_{-\infty}^\infty 2\pi i k \hat{f}(k) e^{2\pi i kx}\,dk. $$ As a result, if $p(t) = \sum_{j=1}^m a_j t^j$ is a polynomial, so that $$ p(D)f(x) := \sum_{j=1}^m a_j D^jf(x) = \sum_{j=1}^m a_j f^{(j)}(x), $$ then one can check that $$ \widehat{p(D)f}(x) = p(2\pi i k)\hat{f}(k), $$ or equivalently, $$ p(D)f(x) = \int_{-\infty}^\infty p(2\pi i k)\hat{f}(k)e^{2\pi i kx}\,dx. $$ On the other hand, it is also a basic property of the Fourier transform that $$ \widehat{S_af}(k) = e^{2\pi i ak}\hat{f}(k) = e^{a(2\pi i k)} \hat{f}(k), $$ or equivalently, that $$ S_af(x) = \int_{-\infty}^\infty e^{a(2\pi i k)}\hat{f}(k)e^{2\pi i kx}\,dk, $$ which suggests, independently of Taylor series, that $S_a = e^{aD}$ in some suitable sense. Hence, if $p$ is any reasonable function defined on (some part of) the imaginary axis of the complex plane, then one can try to define $p(D)$ by $$ p(D)f(x) = \int_{-\infty}^\infty p(2\pi i k)\hat{f}(k)e^{2\pi i k}\,dk $$ for $f$ in some suitable domain---such operators are what one calls pseudo-differential operators.

So, let's apply this machinery to defining the logarithm of $D$. Let $\operatorname{Log}$ denote the principal branch of the logarithm, so that, in particular, $$ \forall k \in \mathbb{R} \setminus \{0\}, \quad \operatorname{Log}(2\pi i k) = \log(2\pi \lvert k \rvert) +i \operatorname{sgn}(k)\frac{\pi}{2}. $$ Then, for any rapidly-decreasing smooth function $f$ such that $$ \lim_{k \to 0} \log(\lvert k \rvert) \hat{f}(k) $$ exists, one should be able to define $$ \operatorname{Log}(D)f(x) = \int_{-\infty}^\infty \operatorname{Log}(2\pi i k)\hat{f}(k)e^{2\pi i kx}\,dk. $$

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Infinitesimal generators for one-parameter groups or semigroups... The literature is huge. You might start here.

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  • $\begingroup$ Such a beautiful subject. $\endgroup$ – Matt Mar 15 '17 at 19:05
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In the Wikipedia entry Calculus of finite differences, you will find references for the classic formulas

$$\triangle f(x) = f(x+1) - f(x) = (e^D-1) \; f(x) \; \; $$ and

$$ D = \ln(1+ \triangle),$$

and their generalization.

In OEIS A238363, you will find several references for interpretations of $\ln(D)$.

For $aD$, just make a change of variables or use $\ln(aD)= \ln(a)+\ln(D)$ for $a>0$.

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