Let's say that I want to find solutions $f\in C(\Bbb R)$ to the equation

$$ f(x+1) + f(x) = g(x) $$

for some $g\in C(\Bbb R)$. I can write $f(x+1) = (Tf)(x)$ where $T$ is the right shift operator and rewrite the equation suggestively as $$ (I+ T)f=g. $$ Formally, I can say that the solution of this equation is

$$ f= (I+ T)^{-1} g. $$

Of course, I am aware that there are infinitely many solutions to the equation but please bear with me for a moment here.

By the theory of operator algebra, if $f,g$ are from some nice Banach space $X$ and our linear operator $T:X\to X$ satisfies $\|T\|<1$, then we have $$ f = \left(\sum_{n=0}^\infty (-T)^n \right)g. $$

However, it is not unreasonable to expect that we should have $\|T\|=1$ for a right-shift operator in most reasonable function spaces so let's try to solve the equation $$ f= (I+\lambda T)^{-1} g. $$ for $\lambda <1$ first then we'll take $\lambda\to 1$. Note that all the steps until now is purely formal since $C(\Bbb R)$ is not a normed space.

For a concrete example, let's say we take $g(x) = (x+2)^2$. The previous method says that we first calculate (for $\lambda<1$) $$\begin{align} f(x) &= \left(I - \lambda T + \lambda^2 T^2 - \dots \right) g(x) \\ &= (x+2)^2 -\lambda (x+3)^1 + \lambda^2 (x+4)^2 + \dots \\ &= \left(1-\lambda+\lambda^2-\dots \right)x^2 + \left(2-3\lambda+4\lambda^2-\dots \right)2x + \left(2^2-3^2\lambda+4^2\lambda^2-\dots \right) \\ &= \frac{1}{1+\lambda} x^2 + 2 \frac{2+\lambda}{(1+\lambda)^2} x + \frac{4+3\lambda + \lambda^2}{(1+\lambda)^3}. \end{align}$$ We shall be brave here and substitute $\lambda=1$ even though the series doesn't converge there. This gives $$ f(x) = \frac 12 x^2 + \frac 32 x + 1 $$ but voilà, for some mysterious reasons unknown to me, this $f$ actually solves our original equation $f(x+1) + f(x) = (x+2)^2$ !

My question is simply:

What are the hidden theories behind the miracle we observe here? How can we justify all these seemingly unjustifiable steps?

I can't give you a reference to this method because I just conjured it up, thinking that it wouldn't work. To my greatest surprise, the answer actually makes sense. I am sure that similar method is probably practiced somewhere, probably by physicists.

Some points worth mentioning:

1.) $C(\Bbb R)$ is probably not the right space to work with since it's not normed. However, I want my answer to be a continuous function on $\Bbb R$ so some form of continuity assumption is needed for our space $X$.

2.) Norming $C(\Bbb R)$ with $L^\infty$ norm is not the way to go since it is possible that $f$ is unbounded, as our example shows.

3.) The solution $f$ is not unique since the kernel of $(I+T)$ consists of all $C(\Bbb R)$ functions $h$ such that $h(x+1)=-h(x)$, e.g. $\sin(\pi x)$.

4.) All the series expansions for $\lambda$ in my example converges when $|\lambda| <1$ but not at $\lambda=1$ but for some reason the result checks out.

  • $\begingroup$ Are you sure about the sign of the 4λ term because putting λ=1 makes the last term 1/4 $\endgroup$ – lux Jun 23 '19 at 22:58
  • $\begingroup$ @lux It should be a plus sign, thanks. $\endgroup$ – BigbearZzz Jun 23 '19 at 22:59
  • 2
    $\begingroup$ While the $\lambda$ series do not converge at $\lambda=1$, they don't diverge there, either. As such, you are using an analytic continuation of the function $\sum_{n=0}^{\infty}T^{n}z^{n}$ (which accepts a complex number and outputs an element of the operator algebra of your Banach space) at $z=-1$. Using holomorphic functional calculus, you can make this analytic continuation argument rigorous. In my experience, when such an analytic continuation exists, the answer you get will be "correct" modulo an arbitrary element of the kernel of the operator you inverted. $\endgroup$ – MCS Jun 24 '19 at 1:25
  • $\begingroup$ @MCS That's a very interesting remark, thanks. I have learned about the holomorphic functional calculus but did not immediately see the connection myself. $\endgroup$ – BigbearZzz Jun 24 '19 at 9:22
  • 2
    $\begingroup$ cross-posted: mathoverflow.net/q/334693/11260 $\endgroup$ – Carlo Beenakker Jun 24 '19 at 15:16

First, if $g(x)$ is an order $n$ polynomial, we can find a unique order $n$ polynomial $f(x)$ that solves your functional equation by equating the polynomial coefficients on both sides of the equation, but on the other hand your solution method is very clever and interesting. For the remainder, consider $f$ and $g$ to be Fourier transformable distributions (test functions that fall off faster than $1/x^n$ should work).

$T$ is a unitary operator with $(T^\dagger f)(x)=f(x-1)$ and eigenvector $e^{ikx}$ has eigenvalue $e^{ik}$, which equals $-1$ for $k=(2n+1)\pi$. Now suppose $g(x)=e^{ikx}$ with $k \neq (2n+1)\pi$, then your series for $(I+\lambda T)^{-1} g$ becomes $(1-\lambda e^{ik}+\lambda^2 e^{i2k}-...)g=(1+\lambda e^{ik})^{-1}g$. Although the series does not converge for $\lambda = 1$ the limit exists. For $k=(2n+1)\pi$ the series becomes $(1+\lambda +\lambda ^2+\lambda ^3+...)g=(1-\lambda )^{-1})g$, which does not have a good limit as $\lambda \mapsto 1$. In fact, if $(I+T)g=0$, then $(I+T)^2f=0$, but for unitary $T$ $(I+T)^2f=0$ does not have any solutions that do not already solve $(I+T)f=0$. The moral of the story is that your solution method fails if the Fourier transform of $g(x)$ is non-zero for any $k=(2n+1)\pi$.

Finally, as to why your series works for polynomials, the Fourier transform of $x^n$ is $(i\frac{d}{dk})^n\delta (k)$, which does not contain any forbidden frequencies.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.