Decay of $C^2$-function where $f,f',f''$ are $\mathscr{L}^1$

Question

If I have a function $f \in C^2(\mathbb{R})$ with $f, f', f'' \in \mathscr{L}^1$, how do I show that $\lim_{\lvert x \rvert \to \infty} f(x) = 0$.

My guess was that I could use Fourier Transformations somehow.

Answer 1

Applying the Fourier transform we get that $\hat{f}(x)$ and $x^2\hat{f}(x)$ are in $L^{\infty}(\mathbb{R})$. Hence $(x^2+1)\hat{f}(x)\in L^{\infty}(\mathbb{R})$, and therefore $$|\hat{f}(x)|\leq C\frac{1}{x^2+1},\qquad x\in \mathbb{R} $$ for some $C>0$. But then $\hat{f}\in L^1(\mathbb{R})$, which entails $f\in C_0(\mathbb{R})$, and in particular $\lim_{|x|\to \infty}f(x)=0$.

It seems like $f'\in L^1$ is not needed.

EDIT: As noted in the comments, this argument is circular, because in order to prove that $x^2\hat{f}(x)=\widehat{f''(x)}$ one needs that $f\in C_0(\mathbb{R})$. It seems to me that one cannot therefore provide a direct solution via the Fourier transform, as defined in $L^1(\mathbb{R})$. Using the Fourier transform on the space of tempered distributions $\mathcal{S}'$, however, we know that the property $$\widehat{u''}=-x^2\hat{u},\qquad u\in \mathcal{S}' $$ always holds, with the operations of differentiation and multiplication by a power as defined in $\mathcal{S}'$. In the special case where $u$ may be identified with a function $f\in L^1(\mathbb{R})\cap C^2(\mathbb{R})$, all the operations considered above (Fourier transform, differentiation, multiplication by a power) agree with the correspondent usual operations as defined for functions, and therefore, we still obtain our result.

I will add an alternative proof, which works under weaker assumptions. Since $f,f'\in L^1$, $f$ belongs to the Sobolev space $W^{1,1}(\mathbb{R})$. It is a known result from basic Sobolev space theory (see e.g. Brezis's book on functional analysis, Corollary 8.9), that for all $f\in W^{1,1}(\mathbb{R})$, we have $$\lim_{|x|\to \infty}f(x)=0 $$ and since $f$ is assumed continuous, we have $f\in C_0(\mathbb{R})$.