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Math Help - Functions of class $C^{1}$?

  1. #1
    Senior Member bkarpuz's Avatar
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    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
    Last edited by bkarpuz; May 2nd 2011 at 09:32 PM.
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  2. #2
    Senior Member bkarpuz's Avatar
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    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}.
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  3. #3
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    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.
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