# Fourier multiplier is the only translation invariant bounded linear operator on $L^2[-\pi, \pi]$

This a question from Stein-Shakarchi Real Analysis.

Let $\mathcal{H}= L^2[ -\pi, \pi]$. And define the $\textbf{Fourier Multiplier}$ by, $$Tf(x) \sim \sum_{-\infty}^{\infty} \lambda_n a_n e^{inx}$$ where $$f(x) \sim \sum_{-\infty}^{\infty} a_ne^{inx}$$

I have shown so far that $T$ commutes with translations, that is $$\tau_h \circ T= T \circ \tau_h, \hspace{3mm} \forall h \in \mathbb{R}$$

I now must show that any bounded linear operator on $L^2[-\pi, \pi]$ that commutes with translation is of the form $T$, the Fourier multiplier.

Let $G$ be another linear bounded operator that commutes with translation, and $f \in L^2[-\pi,\pi]$. Then $$\tau_h \circ G(f) = \tau_h G \left ( \sum_{-\infty}^{\infty} a_n e_n \right)=\tau_h \sum_{-\infty}^{\infty}a_n G(e^{inx})$$

Since $G$ commutes with tranlation we have that $\tau_h \circ G(e^{inx}) = e^{-inh} G(e^{inx})$

I can't finish the proof from here. I notice that $G(e^{inx})$ is an eigenvector of $\tau_h$ but that observation may be fruitless.

• What are $\lambda_n$? – user25959 Feb 3 '15 at 23:10

Let $f(x)=e^{inx}$. Then,
$$\sum_{m= - \infty}^{\infty} \lambda_m e^{im(x-h)} = \tau_h \circ T(f(x)) = T \circ \tau_h (f(x)) = T(e^{in(x-h)}) = e^{-inh} Tf(x) = e^{-inh} \sum_{m= - \infty}^{\infty} \lambda_m e^{imx}$$
Now, $$\sum_{m = - \infty}^{\infty} \lambda_m e^{-imh}e^{imx} = e^{-inh} \sum_{m = -\infty}^{\infty} \lambda_m e^{imx}$$ $$\Longrightarrow \sum_{m= - \infty}^{\infty} \lambda_m e^{imx} \left ( e^{-imh} - e^{-inh} \right ) =0$$
Now for each $m \neq n$, there exists $h \in \mathbb{R}$ such that $e^{-imh} \neq e^{-inh}$, and thus we must have that $\lambda_m = 0$ whenever $m \neq n$.
Thus we must have that $T(e^{inx}) = \lambda_n e^{inx}$.
Now let $f \in L^2 [ - \pi, \pi]$ be arbitrary, we then have that $$T(f(x)) = \sum_{- \infty}^{\infty} a_n \lambda_n e^{inx}$$