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Let $f\in S$ with $S= \left\{ {f:\mathbb{R} \to \left[ {0, + \infty } \right):\int_{ - \infty }^{+\infty} {{{\left| {\hat f\left( t \right)} \right|}^2}{{\left( {1 + {{\left| t \right|}^2}} \right)}^s}dt \leqslant C} } \right\}$, where $a>\frac{1}{2}$, $C>0$, and $\hat f$ is the Fourier transform of $f$ as usual. In a paper which I have read, the authors said that any function $f\in S$ is bounded continuous on $\mathbb{R}$. I am trying to prove this statement. I see that the boundness of $f$ from the class $S$ can be deduced from the fact that $$\left| {f\left( x \right)} \right| = \left| {\frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{itx}}\hat f\left( t \right)dt} } \right| \leqslant \int_{ - \infty }^\infty {\left| {\hat f\left( t \right)} \right|dt} = {\left\| {\hat f} \right\|_{{L^1}\left( \mathbb{R} \right)}}\leqslant const.$$ However, the equality ${f\left( x \right) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{itx}}\hat f\left( t \right)dt} }$ is just hold if $x$ is a continuity point of $f$. Therefore, I focused on proving the continuity of $f$ when $f\in S$. So far I have not found a proof for the statement. My question: How to prove $f$ is continous on $\mathbb{R}$ when $f\in S$? Can someone help me?

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By the way, you have a typo, but I know what you mean: You have an exponent $s$ on $(1+|s|^{2})$, but then you refer to $\alpha > 1/2$. I assume $\alpha=s$ was meant.

If $f=g$ a.e. on $\mathbb{R}$, then $f\in S$ iff $g \in S$, even though one may be unbounded and the other not. What they mean is that there is a continuous function $g$ such that $g=f$ a.e.. If that's not what they mean, then the statement of the problem is wrong because changing $f$ arbitrarily on a set of measure $0$ does not affect the Fourier transform.

You have basically proved the result because $\hat{f} \in L^{1}$ implies $g(t)=\int_{-\infty}^{\infty}e^{itx}\hat{f}(t)\,dt$ is continuous by the Lebesgue dominated convergence theorem, and $g=f$ a.e..

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  • $\begingroup$ In my question, $s>\frac{1}{2}$. I understood your hint. Thank you so much. $\endgroup$
    – Cao
    Sep 18, 2014 at 14:57

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