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Let $fin L^1(mathbb R^d)$. Define the Fourier transform of $f$ by $$hatf(y)=int_mathbb R^df(x)e^-2pi ixcdot y,dx,,,,(yinmathbb R^d).$$
Show that $hatf:mathbb R^dtomathbbC$ is a continuous function and $|hatf|_inftyleq|f|_1.$

$textbfMy Attempt:$ First we prove continuity. Let $yinmathbb R^d$ be arbitrary but fixed, and let $(y_n)_nsubsetmathbb R^d$ such that $y_nto y$ as $ntoinfty$. Then for every $xinmathbb R^d$ we have that $f(x)e^-2pi ixcdot y_nto f(x)e^-2pi ixcdot y$ as $ntoinfty$, since the function $ymapsto e^-2pi ixcdot y$ is continuous. Now observe that $vert f(x)e^-2pi ixcdot y_nvertleqvert f(x)vert$ for all $xinmathbb R^d.$ Therefore by the Dominated convergence theorem, it follows that $$limlimits_ntoinftyhatf(y_n)=int_mathbb R^df(x)e^-2pi ixcdot y,dx=hatf(y).$$ Since $yinmathbb R^d$ was arbitrary, the function $hatf:mathbb R^dtomathbbC$ is continuous.

Now we prove that $|hatf|_inftyleq|f|_1.$ Note that for all $yinmathbb R^d$ we have $$vert hatf(y)vertleqint_mathbb R^dvert f(x)vertvert e^-2pi ixcdot yvert,dx=|f|_1.$$ Since the above holds for all $yinmathbb R^d$, it follows that $|hat f(y)|_inftyleq|f|_1.$

Do you agree with the my proof?

Any feedback is much welcomed. Thank your for your time.

Your proof is correct, but I think it is not the best possible: you use the Dominated convergence theorem, which is quite a strong result, only to prove continuity of the Fourier transform.

Another proof could be based on

$|hatf(y+h)-hatf(y)|=|int_mathbbR^df(x)e^-2pi i xcdot y(e^-2pi ixcdot h-1)|
leqint_mathbbR^d|f(x)||e^-2pi i xcdot h-1|$

This equation one has the advantage not do depend on $y$, and from this we deduce that the continuity is uniform (some work is required to show it thought, since $|e^-2pi i xcdot h-1|$ doesn't go to $0$ uniformly), and we did this without appelling to $hatf$ the Dominated convergence theorem.

For what concerns the boundedness of $hatf$, your proof is perfectly fine.