HALF SOLVED: Fourier inversion formula convergence I have this question in Functional Analysis involving Fourier Inversion formula and the modes of its convergence:

Let $ f \in C^{1}(R)\cap L^1(R)$ with Fourier Transform $ \widetilde{f} $ and define $ f_n(x) = {\frac{1}{2\pi}}\int_{-n}^{n}\widetilde{f}(w)e^{iwx}dw $
  (Fourier Transform : $ \widetilde{f}(w) = \int_{-\infty}^{\infty}f(x)e^{-iwx}dx $)
a. Explain why $ f_n(x)$ is well defined 
MY ANSWER TO A : We know first of all that the Fourier Transform exists for every function in $ L^1(R) $ so $ \widetilde{f} $ exists on all of R and is a continuous function that vanishes at infinity therefore integrable on any interval [-n,n] for all x, and also finite: $ |{\frac{1}{2\pi}}\int_{-n}^{n}\widetilde{f}(w)e^{iwx}dw| \leq {\frac{1}{2\pi}}\int_{-n}^{n}|\widetilde{f}(w)| |e^{iwx}|dw = {\frac{1}{2\pi}}\int_{-n}^{n}\widetilde{f}(w)dw < \infty $ so because Fourier transform is continuous so absolutely integrable on all finite intervals
b. In what sense does the sequence $ f_n(x) $ converge to f(x)?
MY ANSWER TO B: I know the Fourier inversion formula as I learned it we know that $ f \in C^{1}(R)\cap L^1(R) $ so in particular f is absolutely integrable and also it is $ C^1 $ continuous so in particular of course it is piecewise continuous as its first derivative so pointwise inverse convergence occurs so know that $ \lim_{n \to \infty} {\frac{1}{2\pi}}\int_{-n}^{n}\widetilde{f}(w)e^{iwx_0}dw = \frac{f(x_0^+)+f(x_0^-)}{2} $ but because the function f is of course continuous as given then we know it converges pointwise to all x (right and left sides are equal as f is continuous)
c. Is there another sense in which $ f_n $ converges to f ? I HAVE NO IDEA (What else could it be we are dealing with Riemann integrals here so maybe norm convergence in $L^2$? (I know f is not necessarily in $ L^2 $) or something else?)
d. Now assume $ f \in C_C ^ \infty (R) $ (infinitely differentiable with compact support). We must now prove $f_n$ converges to f uniformly. I HAVE NO IDEA

As you can see I have done two of four parts but still cannot do the last two so I need help in solving them. All appreciated.
 A: c. Certainly you cannot guarantee convergence in any $L^p$. You could try convergence in measure and in the sense of distributions.
d. If $f\in C_{C}^\infty(\mathbb{R})$, then $\tilde f\in\mathcal{S}(\mathbb{R})$, the Schwarz class. This implies in particular that $\tilde f(w)$ and $w\,\tilde f(w)$ are in $L^1$. Then
$$
|f_n(x)|\le\frac{1}{2\,\pi}\int_{\mathbb{R}}|\tilde f(w)|\,dw,
$$
that is, the sequence $\{f_n\}$ is uniformly bounded. Moreover, $f_n$ can be differentiated under the integral sign, and
$$
|f_n'(x)|\le\frac{1}{2\,\pi}\int_{\mathbb{R}}|w|\,|\tilde f(w)|\,dw.
$$
By Ascoli-Arzèla's theorem, a subsequence of $\{f_n\}$ converges uniformly, and its limit must be $f$. Since this argument can be applied to any subsequence f $\{f_n\}$, we can conclude that the whole sequence $\{f_n\}$ converges uniformly to $f$.
A: Julian Aguirre has already provided a proof for part d. I offer an alternative one.
Your answers to a. and b. are correct.  As for c. - you were right to guess that $L^2$ had something to do with
the answer. In fact, if you assume that $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ then the sequence will converge to $f$
also in $L^2(\mathbb{R})$. This in fact the content of the inversion formula in $L^1\cap L^2(\mathbb{R})$, and you can find
more details in most texts on Fourier transforms. (e.g. Katznelson, Introduction to Harmonic Analysis, or Rudin, Real and Complex Analysis).
Regarding part d. It is sufficient to assume that $f$ is twice differentiable with a compact support. Since $f$ is $C^2(\mathbb{R})$ and compactly supported, it is in particular in $L^1\cap L^2(\mathbb{R})$, and so the Fourier-inversion formula
is valid for every $x$:
$$f(x)={1\over 2\pi}\int_{-\infty}^{\infty}\tilde{f}(w)e^{iwx}\,dw$$
For every $n$ and $x$: we have:
$$2\pi |f_n(x)-f(x)|=\left|\int_{-n}^n\tilde{f}(w)e^{iwx}\,dw-\int_{-\infty}^{\infty}\tilde{f}(w)e^{iwx}\,dw\right|=\left|\int_{|w|>n}\tilde{f}(w)e^{iwx}\,dw\right|\leq\int_{|w|>n}|\tilde{f}(w)|\,dw$$
By the formula for the Fourier transform of derivatives we have:
$$\tilde{f}(w)={1\over (iw)^2}\widetilde{f''}(w)$$
Since $f$ has compact support, so does it's second derivative, and therefore the Fourier transform of the second derivative is bounded by some absolute constant, say, $C>0$. and so we can estimate the last integral on the r.h.s as:
$$\int_{|w|>n}|\tilde{f}(w)|\,dw\leq C\int_{|w|>n}{1\over w^2}\,dw$$
Thus we have obtained a bound that does not depend on $x$ and tends to zero as $n\to\infty$, because the integral of $1/w^2$ is finite away from the origin.
