# Functions of class $C^{1}$?

• May 2nd 2011, 08:20 PM
bkarpuz
Functions of class $C^{1}$?
Dear MHF members,

the problem is the following.
Show that the functions
$u(x):=\int_{0}^{\infty}\dfrac{\mathrm{e}^{-t} \sin(tx)}{\sqrt{t}}\mathrm{d} t$
and
$v(x):=\int_{0}^{\infty}\dfrac{\mathrm{e}^{-t} \cos(tx)}{\sqrt{t}}\mathrm{d} t$
are of class $C^1$.

Thanks.
bkarpuz
• May 7th 2011, 11:34 AM
bkarpuz
Here I will only show that $u$ is of class $C^{1}$ since the same arguments apply for $v$ too.

Let $x,y\in\mathbb{R}$ and $t\geq0$, then by the mean value theorem, we have $\sin(xt)-\sin(yt)=t\cos(tz_{t})(x-y)$, where $z_{t}$ is between $x$ and $y$, which yields $u(x)-u(y)=\int_{0}^{\infty}\mathrm{e}^{-t}\sqrt{t}\cos(tz_{t})(x-y)\mathrm{d}t.$ Thus, for all $x,y\in\mathbb{R}$, we have $|u(x)-u(y)|\leq\sqrt{\pi}|x-y|/2$ since $\int_{0}^{\infty}\mathrm{e}^{-t}\sqrt{t}\mathrm{d}t=\sqrt{\pi}/2$. Given $\varepsilon>0$, we may let $\delta:=2\varepsilon/\sqrt{\pi}$ to have $|u(x)-u(y)|<\varepsilon$ for all $x,y\in\mathbb{R}$ with $|x-y|<\delta$. This shows that $u$ is continuous.

For the next step, we shall show that for $x\in\mathbb{R}$, the integral $\int_{0}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\sin(tx)\mathrm{d}t$ converges uniformly, i.e., for every $\varepsilon>0$ there exists $r>0$ such that $\bigg|\int_{s}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\sin(tx)\mathrm{d}t\bigg|$ $<\varepsilon$ for all $s\geq r$ and $x\in\mathbb{R}$.

Since $\int_{0}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\mathrm{d}t=\sqrt{\pi}$, for any given $\varepsilon>0$, there exits $r>0$ such that $\int_{r}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\mathrm{d}t<\varepsilon$, which yields $\bigg|\int_{s}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\sin(tx)\mathrm{d}t\bigg|$ $\leq\int_{s}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\mathrm{d}t<\varepsilon$ for all $s\geq r$. Therefore, the integral converges uniformly, which allows us to change the order of the differential operator and the integral. Then, for all $x\in\R$, we have $u^{\prime}(x)=\int_{0}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\big(\sin(tx)\big)^{\prime} \mathrm{d}t$ $=\int_{0}^{\infty}\mathrm{e}^{-t}\sqrt{t}\cos(tx)\mathrm{d}t$ showing that $u^{\prime}$ exists. We can show that $u^{\prime}$ is continuous as well, i.e., $u$ is in $C^{1}$.
• May 7th 2011, 10:44 PM
Jose27
How about this: Call $u_1$ the integrand for $u$, then $|u_1(x,t)| \leq g(t)$ and $\left| \frac{ \partial u_1(x,t)}{\partial x} \right| \leq g_1(t)$ where $g,g_1$ are integrable, and using the mean value theorem and dominated convergence we obtain the result.