# Weak Convergence of Shift Operator on $L^2(\mathbb{R})$

Let $$T_t$$ be the operator $$T_t: \varphi(x) = \varphi(x+t)$$ on $$L^2(\mathbb{R})$$. To what operator does $$T_t$$ converge as $$t \to \infty$$ and in what topology?

I believe that $$T_t$$ should converge weakly to $$T = 0$$. I have outlined my proof below, but I am unsure if the steps are logical (particularly the limits).

For each functional $$\ell \in L^2(\mathbb{R})^*$$, the Riesz Representation Theorem implies that there exists $$\psi \in L^2(\mathbb{R})$$ such that $$\ell(\phi) = (\psi, \phi)$$ for all $$\phi \in L^2(\mathbb{R})$$. Then, \begin{align*} |\ell(T\varphi) - \ell(T_n\varphi)| &= |(\psi, T\varphi) - (\psi, T_n\varphi)| \\ &= |(\psi, 0) - (\psi, T_n\varphi)| \\ &= |(\psi, T_n\varphi)| \\ &= \left|\int_{-\infty}^{\infty} \overline{\psi(x)} \phi(x+n) dx\right| & (\text{Defintion of Inner Product}) \\ &= \left|\int_{-\infty}^{-m} \overline{\psi(x)} \phi(x+n) dx\right| + \left|\int_{-m}^{\infty} \overline{\psi(x)} \phi(x+n) dx\right| \\ &\leq \int_{-\infty}^{-m} \left|\overline{\psi(x)} \phi(x+n)\right| dx + \int_{-m}^{\infty} \left|\overline{\psi(x)} \phi(x+n)\right| dx & (\text{Triangle Inequality}) \\ &\leq \left[\int_{-\infty}^{-m} \left|\overline{\psi(x)} \phi(x+n)\right|^2 dx\right]^{1/2} + \left[\int_{-m}^{\infty} \left|\overline{\psi(x)} \phi(x+n)\right|^2 dx\right]^{1/2} \\ &\leq \left[\int_{-\infty}^{-m} \left|\overline{\psi(x)}\right|^2 \|\phi\|^2 dx\right]^{1/2} + \left[\int_{-m}^{\infty} \|\psi\|^2\left|\phi(x+n)\right|^2 dx\right]^{1/2} & (\text{\psi, \phi bounded}) \\ &= \|\phi\| \left[\int_{-\infty}^{-m} \left|\overline{\psi(x)}\right|^2 dx\right]^{1/2} + \|\psi\| \left[\int_{-m+n}^{\infty} \left|\phi(x)\right|^2 dx\right]^{1/2} \\ &\xrightarrow[n \to \infty]{} \|\phi\| \left[\int_{-\infty}^{-m} \left|\overline{\psi(x)}\right|^2 dx\right]^{1/2} + \|\psi\| \cdot 0 \\ &\xrightarrow[m \to \infty]{} \|\phi\| \cdot 0 + \|\psi\| \cdot 0 \\ &= 0. \end{align*}

• I have decided to rollback the edit to a version with sufficient context (and also because then my answer makes more sense). I also voted to reopen. Jul 5, 2021 at 14:54

I think your guess that $$T_t$$ converges to $$0$$ in some topology (weak operator topology) is correct.
Note: You say that $$T_t$$ converges weakly to $$0$$, but some people call this (what you wanted to show in your proof) weakly operator convergent to $$0$$.
But in my opinion your proof is not correct: you have to show $$\lim_{n\to\infty}|\ell(T\varphi)-\ell(T_n\varphi)|=0$$, but your expression also involves the variable $$m$$ (this variable should be explained somewhere in my opinion). You finish by taking a limit with respect to $$m$$. I do not see how this directly relates to a proof that $$\lim_{n\to\infty}|\ell(T\varphi)-\ell(T_n\varphi)|=0$$.
Maybe you can try to avoid the variable $$m$$ somehow?
In general, if you are in doubt over limits, then you can always go back to the $$\forall \varepsilon>0\exists N\in\Bbb N\dots$$-style definition of the limit.
To be precise, show that, for every $$f\in L^2(\mathbb R)$$, given $$\varepsilon>0$$, there is sufficiently large $$t_o$$ such that, for every $$t\ge t_o$$, $$\Big|\int f(x)\cdot f(x+t)\;dx\Big|<\varepsilon$$. To do so, you'll need to prove that the "tails" of $$f$$ get small, in a precise sense.