Fourier transform extended to $L^2$ Let $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$, and let $f_k$ be functions in the Schwartz class such that $\|f-f_k\|_1+\|f-f_k\|_2\rightarrow 0$ as $k\rightarrow\infty$. Define $$g_k(t)=\int_\mathbb{R}f_k(x)e^{-itx}dx \text{    and    } g(t)=\int_\mathbb{R}f(x)e^{-itx}dx$$ It can be shown that $\lim_{k\rightarrow\infty}g_k(t)=g(t)$ for all $t$.
Let $T_1:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ be defined as the unique continuous mapping that extends the mapping $T:S\rightarrow L^2(\mathbb{R})$, where $S$ is the Schwartz class, and the Fourier transform of a function in the Schwartz class is defined using the $L^1$ definition (like $g$ and $g_k$ above.)
How can I prove that $$\lim_{k\rightarrow\infty}\|g_k-T_1f\|_2=0$$ and also that $$g(t)=(T_1f)(t)$$ pointwise?
EDIT: The first one is a consequence of Plancherel theorem, as Daniel Fischer mentioned in the comment. What about the second one?
 A: The Fourier transform as a mapping $T\colon S \to S$, where $S$ is endowed with the $L^2$-norm, is an isometry (Placherel's theorem). Thus, since $S$ is dense in $L^2(\mathbb{R})$, so is its (unique continuous) extension $T_1 \colon L^2(\mathbb{R}) \to L^2(\mathbb{R})$. Then
$$\lVert g_k - T_1 f\rVert_2 = \lVert T_1 f_k - T_1 f\rVert_2 = \lVert f_k - f\rVert_2 \to 0.$$
The $L^1$ convergence $f_k \to f$ implies uniform convergence $g_k \to g$, and since $L^p$ convergence of a sequence implies the existence of a subsequence that converges pointwise almost everywhere, we have
$$g_k \to T_1 f$$
almost everywhere, hence
$$g = T_1 f$$
almost everywhere. Since $T_1 f$ is only defined as an element of $L^2(\mathbb{R})$, it is only defined modulo functions that are $0$ almost everywhere, so we cannot flatly assert $g(t) = T_1 f(t)$ for all $t$.
However, $g$ is a representative of $T_1 f$, and thus we can say that the most regular representative of the equivalence class $T_1 f$ is the continuous function $g$, in that sense, we have $g(t) = T_1 f(t)$ pointwise.
