Return to Answer

Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$. If we specify those to be odd and with vanishing derivative at the boundaries, i.e., $f(-x)=-f(x)$ and $f^{\prime } (-1/2)=f^{\prime } (1/2)=0$, then $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse: $$ -\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y) $$ If we act with $K$ on an $f$ from our space, we again obtain a result from our space, $$ \int_{-1/2}^{1/2} dy K(x,y) f(y) = \int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) = $$

$$ \int_{-1/2}^{x} dy \left[ y-x \right] f(y) +\int_{x}^{1/2} dy \left[ x-y \right] f(y) = $$

$$ \int_{-x}^{1/2} dy \left[ -x-y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ x+y \right] f(-y) = $$

$$ \int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y) $$ as well as $$ \left. \frac{d}{dx} \int_{-1/2}^{1/2} dy [1-|x-y|] f(y) \right|_{x=1/2} = $$

$$ \left. \int_{-1/2}^{1/2} dy [ 1-2\theta (x-y)] f(y) \right|_{x=1/2} = $$

$$ -\int_{-1/2}^{1/2} dy f(y) =0 $$ Of course, the same is true for acting with the inverse.

This is probably not unique. Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$. If we specify those to be odd and with vanishing derivative at the boundaries, i.e., $f(-x)=-f(x)$ and $f^{\prime } (-1/2)=f^{\prime } (1/2)=0$, then $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse: $$ -\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y) $$ If we act with $K$ on an $f$ from our space, we again obtain a result from our space, $$ \int_{-1/2}^{1/2} dy K(x,y) f(y) = \int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) = $$

$$ \int_{-1/2}^{x} dy \left[ y-x \right] f(y) +\int_{x}^{1/2} dy \left[ x-y \right] f(y) = $$

$$ \int_{-x}^{1/2} dy \left[ -x-y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ x+y \right] f(-y) = $$

$$ \int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y) $$ as well as $$ \left. \frac{d}{dx} \int_{-1/2}^{1/2} dy [1-|x-y|] f(y) \right|_{x=1/2} = $$

$$ \left. \int_{-1/2}^{1/2} dy [ 1-2\theta (x-y)] f(y) \right|_{x=1/2} = $$

$$ -\int_{-1/2}^{1/2} dy f(y) =0 $$ Of course, the same is true for acting with the inverse.

This is probably not unique. Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$, and if we specify those to be odd, $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse: $$ -\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y) $$ If we act with $K$ on an $f$ from our space, we again obtain a result from our space, $$ \int_{-1/2}^{1/2} dy K(x,y) f(y) = \int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) = \int_{-1/2}^{x} dy \left[ y-x \right] f(y) +\int_{x}^{1/2} dy \left[ x-y \right] f(y) =\int_{-x}^{1/2} dy \left[ -x-y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ x+y \right] f(-y) = \int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y) $$ Of course, the same is true for acting with the inverse.

Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$. If we specify those to be odd and with vanishing derivative at the boundaries, i.e., $f(-x)=-f(x)$ and $f^{\prime } (-1/2)=f^{\prime } (1/2)=0$, then $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse: $$ -\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y) $$ If we act with $K$ on an $f$ from our space, we again obtain a result from our space, $$ \int_{-1/2}^{1/2} dy K(x,y) f(y) = \int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) = $$

$$ \int_{-1/2}^{x} dy \left[ y-x \right] f(y) +\int_{x}^{1/2} dy \left[ x-y \right] f(y) = $$

$$ \int_{-x}^{1/2} dy \left[ -x-y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ x+y \right] f(-y) = $$

$$ \int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y) $$ as well as $$ \left. \frac{d}{dx} \int_{-1/2}^{1/2} dy [1-|x-y|] f(y) \right|_{x=1/2} = $$

$$ \left. \int_{-1/2}^{1/2} dy [ 1-2\theta (x-y)] f(y) \right|_{x=1/2} = $$

$$ -\int_{-1/2}^{1/2} dy f(y) =0 $$ Of course, the same is true for acting with the inverse.

This is probably not unique. Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$, and if we specify those to be odd, $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse: $$ -\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y) $$ If we act with $K$ on an $f$ from our space, we again obtain a result from our space, $$ \int_{-1/2}^{1/2} dy K(x,y) f(y) = \int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) = \int_{-1/2}^{x} dy \left[ x-y \right] f(y) +\int_{x}^{1/2} dy \left[ y-x \right] f(y) =\int_{-x}^{1/2} dy \left[ x+y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ -y-x \right] f(-y) = -\int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y) $$ Of course, the same is true for acting with the inverse. HOLD ON THIS ISN'T QUITE RIGHT YET

This is probably not unique. Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$, and if we specify those to be odd, $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse: $$ -\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y) $$ If we act with $K$ on an $f$ from our space, we again obtain a result from our space, $$ \int_{-1/2}^{1/2} dy K(x,y) f(y) = \int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) = \int_{-1/2}^{x} dy \left[ y-x \right] f(y) +\int_{x}^{1/2} dy \left[ x-y \right] f(y) =\int_{-x}^{1/2} dy \left[ -x-y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ x+y \right] f(-y) = \int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y) $$ Of course, the same is true for acting with the inverse.