# Prove $f(x) = \frac 1 {\sqrt{2\pi}} \int_{\mathbb R} \hat f(t) e^{itx} \ \lambda(dt)$ for every $x \in \mathbb R$.

Let $\mathcal L_{\mathbb C}^1(\lambda)$ such that $\hat f \in L_{\mathbb C}^1(\lambda)$ (Fourier transformation).

I've proven that $f(x) = \frac 1 {\sqrt{2\pi}} \int_{\mathbb R} \hat f(t) e^{itx} \ \lambda(dt)$ (this is also equal to the double Fourier transformation of $f(-x)$). $\lambda$-almost-everywhere $x \in \mathbb R$.

Now suppose $f$ is also continuous. Then I want to show the above formula holds for all $x \in \mathbb R$.

I know a result that states that if $f,g$ are two continuous functions then $f = g$ $\lambda$-almost-everywhere $x \in \mathbb R$ $\iff f(x) = g(x)$ for all $x \in \mathbb R$.

But how do I prove $\frac 1 {\sqrt{2\pi}} \int_{\mathbb R} \hat f(t) e^{itx} \ \lambda(dt)$ is continuous ?

• Use the dominated convergence theorem and continuity of $x \mapsto e^{itx}$. – copper.hat Dec 9 '14 at 17:34
• Could you sketch how you would apply it ? – Shuzheng Dec 9 '14 at 17:36
• Let $x \mapsto \phi(x)$ be the right hand side above. If $x_n \to x$ you want to show that $\phi(x_n) \to \phi(x)$. The function $t \mapsto \hat{f}(t) e^{ixt}$ is bounded by the integrable function $t \mapsto |\hat{f}(t)|$, and $\hat{f}(t) e^{ix_nt} \to \hat{f}(t) e^{ixt}$. – copper.hat Dec 9 '14 at 17:57

## 1 Answer

@copper.hat already pointed out how to prove the statement in this particular case, but it is also good to know the following (more general) statement:

Theorem: Let $(X,\mathcal{A},\mu)$ be a measure space and $u: X \times (a,b) \to \mathbb{C}$, $- \infty \leq a < b \leq \infty$, such that

• $x \mapsto u(t,x)$ is continuous for (almost every) $t \in X$
• There exists $w \in L^1$ such that $|u(t,x)| \leq w(t)$ for all $x \in (a,b)$, $t \in X$.

Then the function $V$ defined by $$V(x) := \int_X u(t,x) \, d\mu(t), \qquad x \in (a,b)$$ is continuous.

This follows directly from the dominated convergence theorem (the proof is very similar to the one sketched by @copper.hat in his comment; see e.g. René Schilling: Measures, Integrals and Martingales for a detailed proof).

Here, we have $X=\mathbb{R}$, $(a,b) = (-\infty,\infty)$, $u(t,x) = e^{\imath \ t \cdot x} \hat{f}(x)$. Then $$|u(t,x)| \leq |\hat{f}(t)| =: w(t) \in L^1_{\mathbb{C}}$$ and $t \mapsto u(t,x)$ is (almost everywhere) continuous. Therefore, the claim follows from the above theorem.