finite-difference operator and scalar product in $L_{2}(a,b)$

Question

Let:

$$\Delta_{h}f(x)=\frac{f(x+h)-f(x)}{h}$$

Prove that for any finite function $f\in L_{2}(a,b)$ and any function $g\in L_{2}(a,b)$, following formula holds:

$$\left(\Delta_{h}f,g\right)_{L_{2}(a,b)}=-\left(f,\Delta_{-h}g\right)_{L_{2}(a,b)}$$

Here is my take on it, using the fact that $f(x)$ is finite on $(a,b)$, which means equal $0$ for $x\notin(a,b)$:

\begin{align} (\Delta_{h}f,g)_{L_{2}(a,b)} & = \int_{a}^{b}\frac{f(x+h)-f(x)}{h}\,\overline{g(x)}\,dx \\ & =\int_{a}^{b}\frac{f(x+h)\,\overline{g(x)}}{h}\,dx-\int_{a}^{b}\frac{f(x)\,\overline{g(x)}}{h}\,dx \\ & =\int_{a+h}^{b+h}\frac{f(x)\,\overline{g(x-h)}}{h}\,dx-\int_{a}^{b}\frac{f(x)\,\overline{g(x)}}{h}\,dx \\ & =\int_{a+h}^{b}\frac{f(x)\,\overline{g(x-h)}}{h}\,dx-\int_{a}^{b}\frac{f(x)\,\overline{g(x)}}{h}\,dx \\ & =\int_{a}^{b}\frac{f(x)\,\overline{g(x-h)}}{h}\,dx-\int_{a}^{b}\frac{f(x)\,\overline{g(x)}}{h}\,dx-\int_{a}^{a+h}\frac{f(x)\,\overline{g(x-h)}}{h}\,dx \\ & =\int_{a}^{b}f(x)\,\left(-\frac{\overline{g(x-h)}-\overline{g(x)}}{-h}\right)\,dx-\int_{a}^{a+h}\frac{f(x)\,\overline{g(x-h)}}{h}\,dx \\ & =-\left(f,\Delta_{-h}g\right)_{L_{2}(a,b)}-\int_{a}^{a+h}\frac{f(x)\,\overline{g(x-h)}}{h}\,dx \\ \end{align}

So there is a superfluous term $-\int_{a}^{a+h}\frac{f(x)\,\overline{g(x-h)}}{h}\,dx$, but it doesn't seem possible to eliminate it. Any suggestions here?

Answer 1

Here is my take on it, using the fact that $f(x)$ is finite on $(a,b)$, which means equal $0$ for $x\neq(a,b)$

You can use the same argument for $g(x)$. Rewrite this superfluous term as \begin{equation} -\int_{a-h}^a\frac{f(x+h)\overline{g(x)}}{h}dx \end{equation} $\overline{g(x)}$ is $0$ in the range of integration. So this whole term is $0$.