Show that there exists $C$ constatant such that $||f||_{L^{\infty}(\mathbb{R})}\leq\{ ||f||_{L^2(\mathbb{R}}+||f'||_{L^2(\mathbb{R})} \}, \forall f\in S(\mathbb{R})$.
This is a question in my homework of PDE but I'm unable to to progress beyond a certain point.
MY ATTEMPT (following the professor's hint)
(Fourier inversion) $$ |f(x)|=\left| \frac{1}{2\pi}\int_\mathbb{R}e^{i \langle x,\xi\rangle}\hat{f}(\xi)d\xi \right|\leq \frac{1}{2\pi}\int_\mathbb{R}\left|\hat{f}(\xi)\right|d\xi = \frac{1}{2\pi}\int_\mathbb{R}\left|\hat{f}(\xi)(1+i\xi)\right|\frac{1}{(1+\xi^2)^{1/2}}d\xi\leq $$ (Cauchy-Schwarz inequality) $$ \leq \frac{1}{2\pi}\left(\int_\mathbb{R}\left|\hat{f}(\xi)(1+i\xi)\right|^2d\xi\right)^{1/2}\left(\int_\mathbb{R}\frac{1}{(1+\xi^2)}d\xi\right)^{1/2}=\frac{\sqrt{\pi}}{2\pi}\left(\int_\mathbb{R}\left|\hat{f}(\xi)(1+i\xi)\right|^2d\xi\right)^{1/2}=\frac{\sqrt{\pi}}{2\pi} ||\hat{f}(\xi)(1+i\xi)||_{L^2(\mathbb{R})} $$ from this point I'm unable to work. I don't know how to make the $f$ and $f'$ come on.
Notice that $$\lVert\widehat{f}(\xi)(1+i\xi)\rVert_2\leqslant \lVert \widehat f\rVert_2+ \lVert \xi \cdot \widehat{f(\cdot)}\rVert_2.$$ To conclude, use Plancherel's identity and integrating by part in the definition of the Fourier transform we may decuce a link between $\xi \cdot \widehat{f}$ and the Fourier transform of $f'$.