Given Tf(x), find the equivalent operator m(k)f^(k) in the Fourier transform sense.

Let $f\in L^2(\mathbb R)$, let $$g(x) = Tf(x) = \int^{x+1}_{x} f(s)ds$$ Find $m\in L^\infty(\mathbb R)$ such that $\hat g(k)=m(k) \hat f(k)$.
Use this to show that $T$ is a bounded linear operator on $L^2(\mathbb R)$.

My thoughts so far:

$$\hat g(k) = \int^\infty_{-\infty} Tf(x) e^{-ikx} dx= \int^\infty_{-\infty} \int^{x+1}_{x} f(s)ds e^{-ikx} dx$$ Let $t=s-x$ and therefore $dt=ds$,

$$\hat g(k) = \int^\infty_{-\infty} \int^{1}_{0} f(t+x)dt e^{-ikx} dx$$

We see this is equivalent to $$\hat g(k) = \int^\infty_{-\infty} \int^\infty_{-\infty} \chi_{[0,1]}(t)f(t+x)dt e^{-ikx} dx$$

Letting $t=-y$, $dt=-dy$, then

$$\hat g(k) = \int^\infty_{-\infty}\int^\infty_{-\infty} \chi_{[0,1]}(-y)f(x-y)dy e^{-ikx} dx = \int^\infty_{-\infty}\int^\infty_{-\infty} \chi_{[-1,0]}(y)f(x-y)dy e^{-ikx} dx$$

This is equivalent to

$$\hat g(k) = \int^\infty_{-\infty} (\chi_{[-1,0]}*f)(x) e^{-ikx} dx= \hat{(\chi_{[-1,0]}*f)(x)}$$

Using the properties of the Fourier transform,

$$\hat g(k) = \hat \chi_{[-1,0]}(k) \hat f(k)$$

So then

$$m(k)= \hat \chi_{[-1,0]}(k) = \int^\infty_{-\infty} \chi_{[-1,0]}(x) e^{-ikx} dx = \int^0_{-1} e^{-ikx} dx = [(-ik)^{-1} e^{-ikx} |^0_{-1} ] = i(k)^{-1}(1-e^{ik})$$

However, this $m(k)$ is not in $L^\infty(\mathbb R)$!? Where did I go wrong?

Second part:

If $m(k) \in L^\infty(\mathbb R)$, then

Since the Fourier transform is an isomorphism on $L^2(\mathbb R)$, $$|| Tf(x)|| = || g(x) || = || \hat g(k) || = || m(k) \hat f(k) || \leq || || m(k)||_{L^\infty} \hat f(k) ||\leq || m(k)||_{L^\infty} ||\hat f(k) ||$$

Let $|| m(k)||_{L^\infty} =C$,

$$|| Tf(x)|| \leq C ||\hat f(k) ||= C || f(x) ||$$

Thus $T$ is a bounded operator on $L^2(\mathbb R)$

• Do you know when a function is fourier transformable? Commented Jul 16, 2014 at 9:08
• The Fourier transform is defined for $L^1$ functions, but can be extended to $L^2$ functions. It is also defined (not relevant to this context, perhaps?? for distributions of compact support and slow-growth distributions. Commented Jul 16, 2014 at 17:45

1 Answer

But your function $m$ is $L^\infty(\mathbb R)$, note that \begin{align*} \lim_{k \to 0} \frac{e^{-ik} - 1}k &= \lim_{k\to 0} \left(\frac{\cos k - 1}{k}-i\frac{\sin k}k\right)\\ &= 0 -i = -i \end{align*} So there is some neighbourhood of $0$, where $m$ is bounded, and $m$ is (trivially) bounded on the complement of every neighbourhood of $0$, so $m \in L^\infty(\mathbb R)$.

• I see. I am very uncomfortable with situations like this (is it bounded when it has 1/k?), so I appreciate your explanation. This is funny how I missed this very basic fact. Commented Jul 16, 2014 at 17:28