Let $f\in L^{1}(\mathbb R^d)$. Define the Fourier transform of $f$ by $$\hat{f}(y)=\int_{\mathbb R^d}f(x)e^{-2\pi ix\cdot y}\,dx\,\,\,\,(y\in\mathbb R^d).$$ Show that $\hat{f}:\mathbb R^d\to\mathbb{C}$ is a continuous function and $\|\hat{f}\|_{\infty}\leq\|f\|_{1}.$
$\textbf{My Attempt:}$ First we prove continuity. Let $y\in\mathbb R^d$ be arbitrary but fixed, and let $(y_n)_n\subset\mathbb R^d$ such that $y_n\to y$ as $n\to\infty$. Then for every $x\in\mathbb R^d$ we have that $f(x)e^{-2\pi ix\cdot y_n}\to f(x)e^{-2\pi ix\cdot y}$ as $n\to\infty$, since the function $y\mapsto e^{-2\pi ix\cdot y}$ is continuous. Now observe that $\vert f(x)e^{-2\pi ix\cdot y_n}\vert\leq\vert f(x)\vert$ for all $x\in\mathbb R^d.$ Therefore by the Dominated convergence theorem, it follows that $$\lim\limits_{n\to\infty}\hat{f}(y_n)=\int_{\mathbb R^d}f(x)e^{-2\pi ix\cdot y}\,dx=\hat{f}(y).$$ Since $y\in\mathbb R^d$ was arbitrary, the function $\hat{f}:\mathbb R^d\to\mathbb{C}$ is continuous.
Now we prove that $\|\hat{f}\|_{\infty}\leq\|f\|_{1}.$ Note that for all $y\in\mathbb R^d$ we have $$\vert \hat{f}(y)\vert\leq\int_{\mathbb R^d}\vert f(x)\vert\vert e^{-2\pi ix\cdot y}\vert\,dx=\|f\|_{1}.$$ Since the above holds for all $y\in\mathbb R^d$, it follows that $\|\hat f(y)\|_{\infty}\leq\|f\|_{1}.$
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
$|\hat{f}(y+h)-\hat{f}(y)|=|\int_{\mathbb{R}^d}f(x)e^{-2\pi i x\cdot y}(e^{-2\pi ix\cdot h-1})| \leq\int_{\mathbb{R}^d}|f(x)||e^{-2\pi i x\cdot h}-1|$
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 $|e^{-2\pi i x\cdot h}-1|$ doesn't go to $0$ uniformly), and we did this without appelling to $\hat{f}$ the Dominated convergence theorem.
For what concerns the boundedness of $\hat{f}$, your proof is perfectly fine.