<Home>Qualifying Exams><Analysis>Fall 2013 - Problem 7

Fall 2013 - Problem 7

density argument

Show that there is a dense set of functions f∈L2([0,1])f \in L^2\p{\br{0,1}} such that x↦x−1/2f(x)∈L1([0,1])x \mapsto x^{-1/2}f\p{x} \in L^1\p{\br{0,1}} and ∫01x−1/2f(x) dx=0\int_0^1 x^{-1/2} f\p{x} \,\diff{x} = 0.
Solution.

Since continuous functions are dense in L2([0,1])L^2\p{\br{0,1}}, it suffices to show that we can approximate continuous functions well.

Let g∈C([0,1])g \in C\p{\br{0, 1}} and M=sup⁡ ∣g∣M = \sup\,\abs{g}. For ε>0\epsilon > 0, consider f(x)=Axε−(1/2)f\p{x} = Ax^{\epsilon - \p{1/2}} on [0,δ]\br{0, \delta}, where AA and δ>0\delta > 0 will be chosen later, and f(x)=g(x)f\p{x} = g\p{x} everywhere else. Then

∫01∣f(x)x−1/2∣ dx≤M+∣A∣∫0δx−1/2xε−(1/2) dx=M+∣A∣εδε<∞\int_0^1 \abs{f\p{x}x^{-1/2}} \,\diff{x} \leq M + \abs{A} \int_0^\delta x^{-1/2} x^{\epsilon - \p{1/2}} \,\diff{x} = M + \frac{\abs{A}}{\epsilon} \delta^\epsilon < \infty

as long as ε>0\epsilon > 0, so x−1/2f(x)∈L1([0,1])x^{-1/2}f\p{x} \in L^1\p{\br{0,1}}. Similarly,

∫01∣f(x)∣2 dx≤M2+A2∫0δx2ε−1 dx=M2+A22εδ2ε<∞\int_0^1 \abs{f\p{x}}^2 \,\diff{x} \leq M^2 + A^2 \int_0^\delta x^{2\epsilon-1} \,\diff{x} = M^2 + \frac{A^2}{2\epsilon} \delta^{2\epsilon} < \infty

also. First, we need to ensure that ∫01x−1/2f(x) dx=0\int_0^1 x^{-1/2} f\p{x} \,\diff{x} = 0. We calculate

∫01x−1/2f(x) dx=A∫0δx−1/2xε−(1/2) dx+∫δ1x−1/2g(x) dx≕A∫0δxε−1 dx+I=Aεδε+I.\begin{aligned} \int_0^1 x^{-1/2} f\p{x} \,\diff{x} &= A \int_0^\delta x^{-1/2}x^{\epsilon-\p{1/2}} \,\diff{x} + \int_\delta^1 x^{-1/2} g\p{x} \,\diff{x} \\ &\eqqcolon A\int_0^\delta x^{\epsilon-1} \,\diff{x} + I \\ &= \frac{A}{\epsilon} \delta^\epsilon + I. \end{aligned}

II exists since x−1/2x^{-1/2} is integrable and gg is bounded, and so we may pick

A=−εIδε.A = -\frac{\epsilon I}{\delta^\epsilon}.

Next, we want to pick δ\delta so that ff approximates gg well in L2([0,1])L^2\p{\br{0,1}}:

∥f−g∥L2=∥(f−g)χ[0,δ]∥L2≤∥fχ[0,δ]∥L2+∥gχ[0,δ]∥L2≤(∫0δA2x2ε−1 dx)1/2+Mδ1/2=(A22εδ2ε)1/2+Mδ1/2=(εI22)1/2+Mδ1/2=∣I∣ε1/22+Mδ1/2.\begin{aligned} \norm{f - g}_{L^2} &= \norm{\p{f - g}\chi_{\br{0,\delta}}}_{L^2} \\ &\leq \norm{f\chi_{\br{0,\delta}}}_{L^2} + \norm{g\chi_{\br{0,\delta}}}_{L^2} \\ &\leq \p{\int_0^\delta A^2 x^{2\epsilon-1} \,\diff{x}}^{1/2} + M\delta^{1/2} \\ &= \p{\frac{A^2}{2\epsilon}\delta^{2\epsilon}}^{1/2} + M\delta^{1/2} \\ &= \p{\frac{\epsilon I^2}{2}}^{1/2} + M\delta^{1/2} \\ &= \frac{\abs{I}\epsilon^{1/2}}{\sqrt{2}} + M\delta^{1/2}. \end{aligned}

We may simply take δ=ε\delta = \epsilon to get

∥f−g∥L2≤(∣I∣2+M)ε1/2→ε→00,\norm{f - g}_{L^2} \leq \p{\frac{\abs{I}}{\sqrt{2}} + M}\epsilon^{1/2} \xrightarrow{\epsilon\to0} 0,

which completes the proof.