Consider a linear operator as defined below $$S_N[u(x)] : = \int_\Bbb R \chi_{[-N,N]}(\xi)\hat u(\xi)e^{ix\xi}d\xi, \ u \in L^1(\Bbb R)$$ Prove that $S_n:L^1(\Bbb R) \to C^0_b(\Bbb R)$ is well defined and bounded - $C^0_b(\Bbb R)$ is the space of continuos bounded fuction -
I'm having some trouble understanding exercise requests. First, well definition, the question is to prove that is effectively a linear operator? In this case using integral's propriety we got $$ S_N[\alpha u(x) + \beta v(x)] = \alpha\int_\Bbb R \chi_{[-N,N]}(\xi)\hat u(\xi)e^{ix\xi}d\xi + \beta \int_\Bbb R \chi_{[-N,N]}(\xi)\hat v(\xi)e^{ix\xi}d\xi \\= \alpha S_N[u(x)] + \beta S_N[ v(x)]$$ or instead we have to verify that the image of $u(x)$ via $S_N$is unique? In this case since we have a Fourier transform is everything all right?
For boundness is it enough to show that the norm of the operator is bounded? For example I think that in some sense $$||S_N[u(x)]|| \leq \int_\Bbb R |\hat u(\xi)|d\xi$$ since $u\in L^1(\Bbb R)$ we have a upper bound, altough I couldn't formalize it. Thank you for the help.
I assume that with $\hat u$ you mean the Fourier transform of $u$ and $\chi_{[-N,N]}$ is the characteristic function of the interval $[-N,N]$.
The first point is to show that the operator $S_N$ is well defined, i.e. that it can be applied to elements of its domain $L^1(\mathbb{R})$.
Now since
$$ \hat u (k) = \int_\mathbb{R} \frac{dx}{2\pi} e^{-ikx} u(x) $$
we have \begin{align} \vert \hat u (k)\vert &\le \frac{1}{2\pi} \int_\mathbb{R} \vert u(x) \vert \\ & = \frac{1}{2\pi} \Vert u \Vert_{L^1} \end{align}
Then \begin{align} \left \vert S_N[u](x) \right \vert &\le \int_{-N}^N \vert \hat u (k) \vert dk \\ &\le \frac{2N}{2\pi} \Vert u \Vert_{L^1} \end{align}
Which shows that $S_N$ is well defined. Since $|\hat u (k) |$ is bounded it is integrable on a finite interval, we can use dominated convergence to show that $S_N[u](x)$ is continuous, so it does belong to $C_b^0(\mathbb{R})$.
But also, taking the supremum over $x$
\begin{align} \Vert S_N[u] \Vert_{C^0_b} &= \sup_{x\in\mathbb{R}} \vert S_N[u](x)\vert \\ &\le \frac{2N}{2\pi} \Vert u \Vert_{L^1} \, , \end{align}
so it is a bounded operator.