My question is to eventually prove the following inequality: for $f\in H^1(\mathbb{R})=\{f,f'\in L^2\}$
$$\|f\|_{L^\infty}\leq a\|f\|_{L^2}+\frac{1}{a}\|f'\|_{L^2}, \forall a>0.$$
Here are my thoughts:
1). Sobolev embedding tells us:
$$ H^1(\mathbb{R})\hookrightarrow C^{0,1/2}(\mathbb{R}),$$
for $\Omega$ being a compact subset of $\mathbb{R}$.
2). Coverage in $C^{0,1/2}$ implies Converge in $L^\infty$ (still need compact support?)
3). It seems that we may have now $\|f\|_{L^\infty}\leq C\|f\|_{H^1}$ over any compact subset.
My confusion is how to remove the compactness requirement (Lebesgue Dominated convergence theorem?) and how to prove the final version of that inequality?
New thoughts:
suppose $f_n=f \chi_{[-n.n]}$, then via FTC, we have
$$f_n(z)=\int^z_af'(s)ds+f(a),$$
here we can choose $a$ such that $|f(a)|\leq \frac{1}{2n}\int_{-n}^n|f(s)|ds$, then by Cauchy-Schwartz inequality,
$$|f_n(z)|\leq \sqrt{2n}\|f'\|_{L^2}+\frac{1}{\sqrt{2n}}\|f\|_{L^2}.$$
Then I try to apply DCT, which is not allowed in this case.
Best Answer
Warning your Sobolev embedding is written in the wrong sense.
A way to remove the compact support requirement is using the density of $C^\infty_c$ in $H^1$ (if $f∉ H^1$, the right-hand side of your inequality is $+\infty$ so the inequality is true), but as you remark, you do not get the inequality with the constants that you want.
A way to do it is with the Fourier transform $\hat{f}(y) = \int_{\mathbb{R}} e^{-2i\pi\,xy}f(x)\,\mathrm{d} x$. Using Fourier inversion theorem, the definition of the Fourier transform, multiplying and dividing by $\sqrt{1+\pi^2|x|^2}$ and then using the Cauchy-Schwarz inequality yields $$ \|f\|_{L^∞} = \|\hat{\hat{f}}\|_{L^∞} ≤ \int_{\mathbb{R}}|\hat{f}| ≤ \left(\int_{\mathbb{R}} \frac{\mathrm{d}x}{1+\pi^2|x|^2}\right)^\frac{1}{2} \left(\int_{\mathbb{R}} (1+\pi^2|x|^2)\,|\hat{f}(x)|^2\,\mathrm{d}x\right)^\frac{1}{2} $$ Now we can simplify the right terms by writing $$ \begin{align*} \int_{\mathbb{R}} \frac{\mathrm{d}x}{1+\pi^2|x|^2} &= \frac{1}{\pi}\int_{\mathbb{R}} \frac{\mathrm{d}x}{1+|y|^2} = \left[\frac{\arctan(x)}{\pi}\right]_{x=-\infty}^∞ = 1 \\ \left(\int_{\mathbb{R}} (1+\pi^2|x|^2)\,|\hat{f}(x)|^2\,\mathrm{d}x\right)^\frac{1}{2} &= \left(\int_{\mathbb{R}} |\hat{f}(x)|^2 + \,|\pi x\hat{f}(x)|^2\,\mathrm{d}x\right)^\frac{1}{2} \\ &\leq \left(\int_{\mathbb{R}} (|\hat{f}(x)| + \,|\widehat{\nabla f}(x)|)^2\,\mathrm{d}x\right)^\frac{1}{2} \\ &\leq \left\||\hat{f}| + \,|\widehat{\nabla f}|\right\|_{L^2} \\ &\leq \left\|\hat{f}\right\|_{L^2} + \left\|\widehat{\nabla f}\right\|_{L^2} \\ &\leq \left\|f\right\|_{L^2} + \left\|\nabla f\right\|_{L^2}. \end{align*} $$ This gives the inequality when $a=1$. The possibility to put others $a>0$ comes from the scaling properties. Just apply the above inequality to $f_a(x) := f(x/a^2)$ and remark that $$ \|f_a\|_{L^\infty} = \|f\|_{L^\infty} \\ \|f_a\|_{L^2} = a\,\|f\|_{L^2} \\ \|∇f_a\|_{L^2} = \frac{1}{a}\,\|f\|_{L^2}. $$
I suppose there is perhaps another fastest method without using the Fourier transform and explaining better the fact that we can get the constant $1$ here ?