= Solution
For $f\in L^1(\mathbb R)$, its regular <tempered distribution> acts by integration, and the distributional transform agrees with the function
$$
\widehat f(\lambda)=\int_{\mathbb R}f(x)e^{-i\lambda x}\,dx.
$$
The <triangle inequality> gives the uniform bound
$$
|\widehat f(\lambda)|\leq\int_{\mathbb R}|f(x)|\,dx=\|f\|_1.
$$
If $\lambda_j\to\lambda$, then $f(x)e^{-i\lambda_jx}\to f(x)e^{-i\lambda x}$ pointwise and every integrand is bounded in absolute value by the <Lebesgue integrable function> $|f|$. The <Dominated convergence theorem> therefore proves that $\widehat f$ is <continuous>. In fact the <Riemann-Lebesgue lemma> also gives $\widehat f(\lambda)\to0$ as $|\lambda|\to\infty$.
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