Weak Convergence Of Periodic Functions

 
Let $g_j(x)=g(j x), x ∈(0,1)$, where $g$ is $T$-periodic and on $(0, T]$ is given by \[ g=-117\mathit1_{\left(0, \frac{T}{2}\right]}+117\mathit1_{\left(\frac{T}{2}, T\right]}= \begin{cases}-117 & \text { on }\left(0, \frac{T}{2}\right] \\ +117 & \text { on }\left(\frac{T}{2}, T\right]\end{cases} \] Clearly $\left\|g_j\right\|_1=117 \not → 0$. On Problem Sheet 3 you will be asked to prove that $g_j → 0$ in $𝒟'(0,1)$. Problem 3. Let $g ∈ \mathrm{L}_{\text{loc}}^1(ℝ)$ and assume that $g$ is $T$ periodic for some $T>0: g(x+T)=g(x)$ holds for almost all $x ∈ ℝ$. Define for each $j ∈ ℕ$ the function \[ g_j(x)=g(j x),   x ∈(0,1) . \] Prove that \[ g_j → \frac{1}{T} ∫_0^T g \mathrm{~d} x   \text { in }   𝒟'(0,1)   \text { as }   j → ∞ . \] Proof. (page 11) By density assume $g$ is piecewise continuous. If $\phi \in 𝒟(0,1)$, define \[ G(x)=\int_0^x g(t) d t,  x \in(0, T], \] Replacing $g$ with $g-\frac1T∫_0^T g \mathrm{~d} x$, we can assume that $G(T)=0$, so the $T$-periodic extension of $G$ is piecewise $C^1$ and $G'(x)=g(x)$ a.e. Integrate by parts, using $\phi(0)=\phi(1)=0$, \begin{aligned} \left\langle g_j, \phi\right\rangle&=\int_0^1 g(j x) \phi(x) d x\\ &={\left[\frac{G(j x)}{j} \phi(x)\right]_0^1-\frac1j\int_0^1 G(j x)\phi'(x) d x} \\ & \rightarrow 0 . \end{aligned}