Suppose we have a function $\psi$ in $\mathbb{R}$ other than a polynomial which is equal to its Taylor Series for every point in $\mathbb{R}$. Why is the following statement valid?
When we interpret both the $n$th sum of the Taylor series $T_n$ as a tempered distribution and $\psi$ as a tempered distribution, $\psi$ is NEVER the pointwise limit of $T_n$.
The only guess I have is it is because each $T_n$ is not a tempered distribution? I am new to tempered distributions so please explain in detail not assuming much prior knowledge.
Edit: I also know that a distribution is tempered iff it is a finite sum of derivatives of continuous functions growing at infinity slower than some polynomial. So maybe that can help?
THANKS!
The trouble is much more obvious when one looks at the sequence of Fourier transforms $\widehat{T}_n$ of your sequence $T_n$. Let me be precise. Suppose that $\{a_k\}$ is the sequence of Taylor coefficients for $\psi$, where $\psi$ is a real analytic function that is not a polynomial, so that $\psi(x) = \sum_{k\geq 0}a_kx^k$ and $a_k\not = 0$ for infinitely many $k$. Following your notation, write $T_n$ for the $n$th partial sum of this power series. The Fourier transform of $T_n$ is going to be $$ \widehat{T}_n(x) = \sum_{k = 0}^n a_ki^k\partial^k\delta_0, $$ since the Fourier transform of $x^k$ is $i^k\partial^k \delta_0$ (depending on the normalization you use for the Fourier transform this formula may be off by a constant factor, but this is irrelevant); here $\partial$ just denotes differentiation and $\delta_0$ is the standard Dirac delta function at the origin. Now, if you assume that $T_n \to \psi$ in the sense of distributions, then you must also assume that $\widehat{T}_n \to \hat{\psi}$ in the sense of distributions; in particular, you must assume that $$ \lim_{n\to \infty} \widehat{T}_n $$ exists as a tempered distribution. And this is where things fall apart. If we choose a function $\varphi$ in the Schwartz space, then $\partial^{(k)}\delta_0(\varphi) = (-1)^k\varphi^{(k)}(0)$ and $$ \widehat{T}_n(\varphi) = \sum_{k = 0}^n a_ki^k\partial^k\delta_0(\varphi) = \sum_{k = 0}^n a_k (-i)^k\varphi^{(k)}(0). $$ So if $T_n(\varphi)$ is to converge to a finite limit for every $\varphi$, then it must be the case that $$ \sum_{k = 0}^\infty a_k (-i)^k\varphi^{(k)}(0) \tag{1} $$ is a convergent series for every Schwarz function $\varphi$. One way to see that this can never be the case is to invoke Borel's theorem, which ensures that we can find a Schwartz function $\varphi$ such that, for instance, $$ \varphi^{(k)}(0) = \left\{\begin{array}{ll} \frac{1}{(-i)^ka_k} & \text{if $a_k\not = 0$,}\\ 0 & \text{if $a_k = 0$.} \end{array} \right. $$ Then since infinitely many of the $a_k$ are nonzero by assumption, inserting this $\varphi$ into $(1)$ gives us a divergent series.
It may be worth mentioning, however, that whenever $\varphi$ is compactly supported, we do in fact get the convergence $\lim_{n\to \infty} T_n(\varphi) = \psi(\varphi)$; actually, if you take $\psi = e^{-itx}$, you'll be able to show with an argument similar to the one above that the Fourier transform $\hat{\varphi}$ must be real analytic (i.e., that it has an extension to an entire function on the complex plane).