Given $f\in L^2(\mathbb R)$ and $g(x)=\int^{x+1}_x f(t)\,dt$
(a) Relate $\hat f$ and $\hat g$
(b) Prove $g\in H^1(\mathbb R)$
Part A:
$$\hat g(k)= \int^{\infty}_{-\infty} \int^{x+1}_{x} f(t)\,dt e^{-ikx}\,dx $$
Using integration by parts,
$$\hat g(k)= \lim_{b\to\infty }\left [\int^{x+1}_x f(t)\,dt e^{-ikx}(-ik)^{-1} \right ]^b_{-b}- \int^{\infty}_{-\infty} (f(x+1)-f(x))e^{-ikx}(-ik)^{-1}\,dx $$
I'm assuming the boundary term vanishes since $f\in L^2(\mathbb R)$. Continuing,
$$\hat g(k)=i(k)^{-1}\int^{\infty}_{-\infty}f(x)e^{-ikx}\,dx -i(k)^{-1}\int^{\infty}_{-\infty}f(x+1)e^{-ikx}\,dx $$
Using $x+1=y$,
$$\hat g(k)=i(k)^{-1}\int^{\infty}_{-\infty}f(x)e^{-ikx}dx -i(k)^{-1}\int^{\infty}_{-\infty}f(y)e^{-ik(y-1)}\,dx $$
So
$$\hat g(k)=\frac{i}{k}\hat f(k) -\frac{ie^{ik}}{k}\hat f(k) $$
Part B:
$$ \|g\|_{H^1}=\|g\|_{L^2}+\|g'\|_{L^2}=\|\hat g\|_{L^2}+\|f(x+1)-f(x)\|_{L^2} $$
We see that
$$\|\hat g\|_{L^2}= \int^\infty_{-\infty} \left | \frac{i}{k}\hat f(k) -\frac{ie^{ik}}{k}\hat f(k) \right|^2 dk = \int^\infty_{-\infty} \frac{\left |1-e^{ik}\right |^2}{k^2} \left | f(k) \right|^2 \, dk $$
Using the Holder inequality, we see that $ \frac{\left |1-e^{ik}\right |^2}{k^2}$ is bounded by say $M$ and so
$$\|\hat g\|_{L^2} \leq M \|\hat f\|_{L^2(\mathbb R)}= M \| f\|_{L^2(\mathbb R)} $$
We also see that
$$\|f(x+1)-f(x)\|_{L^2} \leq \|f(x+1)\|_{L^2} +\|f(x)\|_{L^2} = 2\|f(x)\|_{L^2}$$
So
$$ \|g\|_{H^1} \leq (2+M)\|f(x)\|_{L^2} $$
Thus $g\in H^1 $
So my question is: Did I handle the integration by parts correctly? Also, do you agree with the rest of this?