Find a function in $H^{\frac{1}{2}}$ that is not in $L^{\infty}$. Let $\mathcal{S}$ be the Schwartz class and $\mathcal{S}'$ be its dual (also known as the set of tempered distributions). Let $\hat{u}$ denote the Fourier transform of the function $u$. Given a real number $s$, define the Sobolev space $H^{s}(\mathbb{R}) = \left\{u\in\mathcal{S}'(\mathbb{R}) : (1+\lambda^{2})^{\frac{s}{2}}\hat{u}(\lambda)\in L^{2}(\mathbb{R})\right\}$. 
I'm trying to solve the following problem: can you find a function in $H^{\frac{1}{2}}(\mathbb{R})$ that is not in $L^{\infty}(\mathbb{R})$?
The problem gives a hint, but I think the hint was not written correctly. The hint is: choose a non-negative function $f\in L^{2}(\mathbb{R})$ such that $(1+\lambda^{2})^{\frac{-1}{4}}f(\lambda)\notin L^{1}(\mathbb{R})$. Then consider $u$ as the inverse Fourier transform of $(1+\lambda^{2})^{\frac{-1}{4}}f(\lambda)$. 
Does anyone know if this hint is correct, or if there is another way to solve the problem?
 A: The hint is correct. The (inverse) Fourier transform of a nonnegative non-$L^1$ function $f$ blows up at $0$. A way to show it rigorously is to multiply $f$ by a "wide" Gaussian $e^{-\delta \xi^2}$ thus obtaining an $L^1$ function with large $L^1$ norm. The inverse Fourier transform of this product is the convolution of $\check{f}$ with a "narrow" Gaussian of unit $L^1$ norm. Direct evaluation shows that the convolution has a  large value at $0$. This gives a lower bound for $\|\check f\|_{L^\infty}$. 
An explicit example can be given: $\log\log |x|^{-1}$ or $\log^{p} |x|$ for $0<p<1$ (I mean to express the behavior near $0$ with these formulas, not the global behavior). A verification would be more difficult, though. If one knows that $H^{1/2}$ norm on a line is equivalent to $H^1$ norm of the harmonic extension to halfplane/disk,  then it's easy: observe that the obvious extension $\log\log |z|^{-1}$ or $\log^{p} |z|$  has finite $H^1$ norm, and the harmonic extension minimizes $H^1$ norm.
