Estimate of a Fourier Multiplier Operator Let $m_t (\xi) = \cos (2\pi |\xi| t).$ Define the operators, for $t>0,$
$$ T_t f = ( m_t \widehat{f} )^{\vee}.$$
It is asked to prove that, whenever $f$ is sufficiently regular, 
$$ \| T_t f\|_{\infty} \le Ct^{-1} \| \nabla f \|_{1,2},$$
Where 
$$\| g \|_{1,2}^2 = \int |\widehat{g}(\xi)|^2 (1+|\xi|^2) d\xi.$$ 
My Attempt
It was a pretty simple strategy: changing variables to obtain certain decay on $t$, writing 
$$1 = \frac{(1+|\xi|^2)|\xi|}{(1+|\xi|^2)|\xi|}$$ 
(Or something like it) and using Cauchy-Schwarz. Unfortunately, with these I only get the desired bound without the $1/t$ factor.
If anyone can improve this and solve the problem, I would be really thankful.
EDIT
I think I have committed a terrible mistake; the $\| \cdot \|_{1,2}$ should be taken to be $\|Df\|_1+ \|D^2 f\|_{1}$. In this case, the scaling argument shows us that the order of decay is right, as $t \rightarrow \infty$. Moreover, by scaling back, we may show only that 
$$ \| T_1 f\|_{\infty} \le C \| \nabla f \|_{1,2}.$$
 A: So we're in dimension $n=3$ according to the problem statement. I agree with you that we can get the inequality without the $1/t$ factor basically by Cauchy-Schwarz. But am I being silly or does the stated inequality not hold by scaling? Suppose the inequality were true. Let $f_{t}=f(\cdot/t)$. Then
$$|T_{t}(f_{t^{-1}})(x)|=|\int_{\mathbb{R}^{3}}\cos(2\pi|\xi| t)t^{3}\widehat{f}(t\xi)e^{2\pi ix\cdot \xi}d\xi|=|\int_{\mathbb{R}^{3}}\cos(2\pi|\xi|)\widehat{f}(\xi)e^{2\pi ix\cdot\xi/t}d\xi|$$
Since the $L^{\infty}$ is unchanged by dilations. We see that
$$\|T_{t}(f_{t^{-1}})\|_{L^{\infty}}=\|T_{1}(f)\|_{L^{\infty}}$$
It is well known that your Sobolev norm is equivalent (with dimensional constants) to the Sobolev norm $\|g\|_{L^{2}}+\|\nabla g\|_{L^{2}}$. Observe that
$$\|\nabla(f_{t^{-1}})\|_{L^{2}}=(\int_{\mathbb{R}^{3}}t^{-2}|(\nabla f)(x/t)|^{2}dx)^{1/2}=t^{1/2}\|\nabla f\|_{L^{2}}$$
and
$$\|\nabla^{2}(f_{t^{-1}})\|_{L^{2}}=(\int_{\mathbb{R}^{3}}t^{-4}|(\nabla^{2}f)(x/t)|^{2}dx)^{1/2}=t^{-1/2}\|\nabla^{2}f\|_{L^{2}}$$
So assuming the inequality were true, we would then have
$$\|T_{1}(f)\|_{L^{\infty}}\lesssim t^{-1/2}\|\nabla f\|_{L^{2}}+t^{-3/2}\|\nabla^{2}f\|_{L^{2}}$$
Taking $f$ to be a suitable nonzero Schwartz function would yield a contradiction as $t\rightarrow\infty$.

Here's my proof of the original inequality without the factor of $1/t$. Observe by triangle inequality
$$|T_{t}(f)(x)|\leq \int_{|\xi|\leq 1}|\widehat{f}(\xi)|d\xi+\int_{|\xi|>1}|\widehat{f}(\xi)|d\xi$$
By Cauchy-Schwarz and Plancherel,
\begin{align*}
\int_{|\xi|\leq 1}\frac{|\xi|}{|\xi|}|\widehat{f}(\xi)|d\xi&\leq (\int_{|\xi|\leq 1}|\xi|^{-2}d\xi)^{1/2}(\int_{\mathbb{R}^{3}}|\xi|^{2}|\widehat{f}(\xi)|^{2}d\xi)^{1/2}\\
&\lesssim\|\nabla f\|_{L^{2}}
\end{align*}
Similarly,
\begin{align*}
\int_{|\xi|>1}\frac{|\xi|^{2}}{|\xi|^{2}}|\widehat{f}(\xi)|d\xi&\leq(\int_{|\xi|>1}|\xi|^{-4}d\xi)^{1/2}(\int_{\mathbb{R}^{3}}|\xi|^{4}|\widehat{f}(\xi)|^{2}d\xi)^{1/2}\\
&\lesssim\|\Delta f\|_{L^{2}}\\
&\lesssim\|\nabla^{2}f\|_{L^{2}}
\end{align*}
A: Even though Matt has solved the original problem properly, I feel obligated to give a proper solution to the fixed version. This is just for completeness:
Step 1: By scaling, we may suppose that $t=1$. This is a calculation just like the one done by Matt above.
Step 2: By translating $f$ if necessary, we may restrict ourselves to the evaluation of $|T_1 f(0)|.$ This is so because of the nice properties the Fourier Transform has under translations.
Step 3: We split $2\cos(2\pi |\xi|) = e^{2\pi i |\xi|} + e^{-2\pi i |\xi|}$ and calculate each part separately:
$$ \int e^{2\pi i |\xi|} \widehat{f}(\xi) d\xi \\ = C \lim_{R \to \infty} \int_{B(0,R)} \frac{e^{2\pi i |\xi|}}{|\xi|^2} \widehat{\Delta f}(\xi) d\xi $$
$$ =  C \lim_{R \to \infty} \int (\int_{B(0,R)} e^{-2\pi i x\xi} e^{2\pi i |\xi|} \frac{1}{|\xi|^2} d\xi ) \Delta f(x) dx,$$
Where we have used Fubini's Theorem. So, in order to achieve the desired bound, it suffices to estimate
$$\int_{B(0,R)} e^{-2\pi i x\xi} e^{2\pi i |\xi|} \frac{1}{|\xi|^2} d\xi.$$
But this is standard: a use of polar coordinates formulas shows that, after some boresome calculations, this is of order at most $\frac{\log(1+|x|)}{|x|},$ and this is known to be bounded. 
The result then follows by applying the same procedure to the other term resulting in the splitting of cosines.
