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where $f'',f',f \to 0$ as $ |x| \to \infty $ and $ g \to 0 \ $ as $|x| \to \infty $ ?

I do not understand how I would add $f''+f$ when one of them would be an integral and the other a function of $x$ ? I am also unsure of how I would get $f''$ since $f$ is a function of x and y.

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    @MyGlasses y cannot be a constant – 2017-01-14

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With a fixed $x$, we can change limits of integral: \begin{eqnarray} 2f(x)&=&\int_{-\infty}^{\infty}e^{-|x-y|}g(y)\,dy\\ &=& \int_{-\infty}^{x}e^{-|x-y|}g(y)\,dy+\int_{x}^{\infty}e^{-|x-y|}g(y)\,dy \\ &=& \int_{-\infty}^{x}e^{-x+y}g(y)\,dy+\int_{x}^{\infty}e^{x-y}g(y)\,dy\\ &=&e^{-x}\int_{-\infty}^{x}e^{y}g(y)\,dy+e^{x}\int_{x}^{\infty}e^{-y}g(y)\,dy\\ &=& I+J \end{eqnarray} Where $$I=e^{-x}\int_{-\infty}^{x}e^{y}g(y)\,dy~~~~,~~~~J=e^{x}\int_{x}^{\infty}e^{-y}g(y)\,dy$$ we know from $\displaystyle\lim_{x\to\pm\infty}g(x)=0$ that \begin{eqnarray} I'&=&-e^{-x}\int_{-\infty}^{x}e^{y}g(y)\,dy+e^{-x}\Big[e^{x}g(x)-\lim_{x\to-\infty}e^{x}g(x)\Big]\\ &=&-e^{-x}\int_{-\infty}^{x}e^{y}g(y)\,dy+g(x)\\ &=&-I+g(x) \end{eqnarray} \begin{eqnarray} J'&=&e^{x}\int_{x}^{\infty}e^{-y}g(y)\,dy+e^{x}\Big[\lim_{x\to+\infty}e^{-x}g(x)-e^{-x}g(x)\Big]\\ &=&e^{x}\int_{x}^{\infty}e^{-y}g(y)\,dy-g(x)\\ &=&J-g(x) \end{eqnarray} S \begin{eqnarray} 2f(x)&=&I+J\\ 2f'(x)&=&-I+J\\ 2f''(x)&=&I-g(x)+J-g(x)\\ &=&2f(x)-2g(x)\\ -f''(x)+f(x)&=&g(x) \end{eqnarray} as we desired.