# Find a relationship between $\hat f$ and $\hat g$, given $g(x)=\int^{x+1}_x f(t)\,dt$, $f\in L^2(\mathbb R)$,

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?

• Alternatively, you could observe that $g=f* h$, with $h=\chi_{(-1,0)}$, and take it from there. – user138530 Aug 17 '14 at 22:36
• Or $g'(x) = f(x+1) - f(x)$, apply the Fourier transform and use properties to relate $\hat{f}$ and $\hat{g}$. – Mark Fantini Aug 17 '14 at 23:01
• There is a problem with the first equation. g is in $L_2$, not necessarily $L_1$, so the integral form of the Fourier transform may not be valid. – Dunham Oct 24 '14 at 12:03