# 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
$\displaystyle u(x):=\int_{0}^{\infty}\dfrac{\mathrm{e}^{-t} \sin(tx)}{\sqrt{t}}\mathrm{d} t$
and
$\displaystyle v(x):=\int_{0}^{\infty}\dfrac{\mathrm{e}^{-t} \cos(tx)}{\sqrt{t}}\mathrm{d} t$
are of class $\displaystyle C^1$.

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

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

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