# Why is $\int_{-\infty}^{\infty} f(x)\sin(tx)dx$ continuous?

This is embarrassing.

I asked this question several months ago:

Let $f$, a Lebesgue integrable function in $\mathbb{R}$ ($\int_{\mathbb{R}}|f| < \infty$). Let: $$g(t) := \int_{-\infty}^{\infty} f(x)\sin(tx)dx.$$ Show that $g$ is continuous and that: $$\lim_{|t| \rightarrow \infty} g(t) = 0.$$

I wrote there that I proved that $g$ is continuous, but I can't for life of me remember my proof, come up with a new one or find my notes. Help?

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Show that for each $t$ and each sequence $\{t_n\}$ which converges to $t$ we have $g(t_n)\to g(t)$. To see this, define $F_n(x):=f(x)\sin(t_nx)$ and $F(x)=f(x)\sin(tx)$.

• By continuity of $y\mapsto \sin y$, we have $F_n(x)\to F(x)$ for each $x$.
• We have $|F_n(x)|\leq |f(x)|$ and $f$ is integrable.

Hence we can apply dominated convergence theorem.

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It's actually the imaginary part of the Fourier transform of $f$. – Davide Giraudo Aug 5 '12 at 12:37
The fact that the limit is zero is the Riemann-Lesbegue lemma. – Potato Dec 7 '12 at 22:11