Proof of Fourier Inverse formula for $L^1$ case I know this may be a stupid question, but still hope someone can help me. I was trying to prove the Fourier inversion formula for which $f$ and $\hat{f}=\int_{\mathbb{R}}f(x)e^{-i2\pi xy}dx$ both lie in $L^1$, and then $f=\int_{\mathbb{R}}\hat{f}(y)e^{i2\pi xy}dy$. I assume the knowing of FT of the Gaussian function $g_c(x)=e^{-c\pi x^2}$, which is $\frac{1}{\sqrt{c}}e^{-\frac{1}{c}\pi x^2}$. Then I have
$$
\int_{\mathbb{R}}\hat{f}(y)e^{i2\pi xy}dy=\lim_{c\to0}\int_{\mathbb{R}}\hat{f}(y)g_c(y)e^{i2\pi xy}dy\\
$$
Using the definition of $\hat{f}$ and Fubini, this becomes
$$
\lim_{c\to0}\int_{\mathbb{R}}\int_{\mathbb{R}}f(t)e^{-i2\pi ty}dtg_c(y)e^{i2\pi xy}dy=
\lim_{c\to0}\int_{\mathbb{R}}f(t)\hat{g}_c(t-x)dt
$$ 
Now I came to the problem. Formally one can guess that $\hat{g}_c(t)$ is just the Dirac function as $c\to0$ so that convolution gives $f$ itself, but how can one show that strictly, especially for an $L^1$ function? On the other hand, I also directly calculated $g_c(t-x)$ which gives me
$$
\lim_{c\to0}\int_{\mathbb{R}}f(t)\frac{1}{\sqrt{c}}e^{-\frac{1}{c}\pi(t-x)^2}dt. $$Change of variables gives me
$$
\lim_{c\to0}\int_{\mathbb{R}}f(x-\sqrt{c}y)e^{-\pi y^2}dy
$$
Then why is it okay for me to pass the limit directly to $f$, since $f$ is just absolutely integrable instead of being continuous? I know this seems true formally, but, again, how can one show this strictly? Thank you for help!                     
 A: First of all, the equality $f(x)=\int_{\mathbb{R}}\hat{f}(y)e^{i2\pi xy}\,dy$ has to be understood properly. Changing $f$ on a set of measure zero does not change $\hat f$. One can says that the equality holds a.e. Or that both sides define the same element of $L^1$. Or that $f$ has a bounded continuous representative, and then the equality holds everywhere for that representative. 
Back to your question. Let $\phi(x)=\exp(-\pi x^2)$. All we need to know about this function is that $\phi\in L^1(\mathbb R)$ and $\int_{\mathbb R} \phi=1$. For $b>0$, let $\phi_b(x)=b^{-1}\phi(b^{-1}x)$. (Here, $b$ corresponds to $\sqrt{c}$ in your notation.)  
Fact. For every $f\in L^1(\mathbb R)$, $f*\phi_b\to f$ in $L^1$. 
Assuming the Fact, you can conclude  that $f$ agrees with the inverse Fourier transform of $\hat f$ in the sense of $L^1$. The latter  function is bounded and continuous, which opens the door to other interpretations stated above.
The Fact is standard:
$$f*\phi_b(x)-f(x)=\int_{\mathbb R}  (f(x-t)-f(x))\phi_b(t)\,dt = \int_{\mathbb R}  (f(x-bs)-f(x))\phi(s)\,ds $$
The $L^1$ norm of the function $x\mapsto f(x-bs)-f(x)$ is bounded by $2\|f\|_{L^1}$. This norm also tends to zero as $b\to 0$, by continuity of translation in $L^1$. The integral Minkowski inequality and dominated convergence theorem finish the proof:
$$\|f*\phi_b-f\|_{L^1}\le \int_{\mathbb R}  \|f(x-bs)-f(x)\|_{L^1} \phi(s)\,ds \to 0 $$
