<Home>Qualifying Exams><Analysis>Fall 2009 - Problem 2

Fall 2009 - Problem 2

Fourier analysis, Laplacian, Lp spaces

Let vv be a trigonometric polynomial in two variables, that is,

v(x,y)=∑n,m∈Zan,me2πi(nx+my),v\p{x, y} = \sum_{n,m\in\Z} a_{n,m}e^{2\pi i\p{nx + my}},

with only finitely many non-zero coefficients an,ma_{n,m}. If

u=v−Δvu = v - \Delta v

where Δ=∂x2+∂y2\Delta = \partial_x^2 + \partial_y^2 is the Laplacian, prove that

∥v∥L∞([0,1]×[0,1])≤C∥u∥L2([0,1]×[0,1])\norm{v}_{L^\infty\p{\br{0,1}\times\br{0,1}}} \leq C \norm{u}_{L^2\p{\br{0,1}\times\br{0,1}}}

for some constant CC independent of vv.
Solution.

From a straightforward calculation,

u(x,y)=∑n,m∈Z(1−4π2(n2+m2))an,me2πi(nx+my),u\p{x, y} = \sum_{n,m\in\Z} \p{1 - 4\pi^2\p{n^2 + m^2}} a_{n,m} e^{2\pi i\p{nx+my}},

so by orthonormality of the e2πi(nx+my)e^{2\pi i\p{nx+my}}, we have

∥u∥L2([0,1]×[0,1])2=∑n,m∈Z(1−4π2(n2+m2))2∣an,m∣2.\norm{u}_{L^2\p{\br{0,1}\times\br{0,1}}}^2 = \sum_{n,m\in\Z} \p{1 - 4\pi^2\p{n^2 + m^2}}^2 \abs{a_{n,m}}^2.

On the other hand, by Cauchy-Schwarz,

∣v∣2≤(∑n,m∈Z∣an,m∣)2=(∑n,m∈Z11−4π2(n2+m2)⋅(1−4π2(n2+m2))∣an,m∣)2≤(∑n,m∈Z1(1−4π2(n2+m2))2)∥u∥L2([0,1]×[0,1])2≕C2∥u∥L2([0,1]×[0,1])2\begin{aligned} \abs{v}^2 \leq \p{\sum_{n,m\in\Z} \abs{a_{n,m}}}^2 &= \p{\sum_{n,m\in\Z} \frac{1}{1 - 4\pi^2\p{n^2 + m^2}} \cdot \p{1 - 4\pi^2\p{n^2 + m^2}}\abs{a_{n,m}} }^2 \\ &\leq \p{\sum_{n,m\in\Z} \frac{1}{\p{1 - 4\pi^2\p{n^2 + m^2}}^2}} \norm{u}_{L^2\p{\br{0,1}\times\br{0,1}}}^2 \\ &\eqqcolon C^2 \norm{u}_{L^2\p{\br{0,1}\times\br{0,1}}}^2 \end{aligned}

The series defined as C2C^2 converges, and so taking square roots, we get

∥v∥L∞([0,1]×[0,1])≤C∥u∥L2([0,1]×[0,1]).\norm{v}_{L^\infty\p{\br{0,1}\times\br{0,1}}} \leq C \norm{u}_{L^2\p{\br{0,1}\times\br{0,1}}}.