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}