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Theorem 2.6. Let $L/K$ be a finite Galois extension and $F/K$ be an arbitrary field extension. The extension $LF/F$ is finite Galois and $\DeclareMathOperator{\Gal}{Gal}\Gal(LF/F)≅\Gal(L/L∩F)$ by restriction. In particular, $[LF:F]=[L:L∩F]$ Proof. Since $L/K$ is finite Galois, $L$ is a splitting field over $K$ of a separable polynomial $f(X) ∈ K[X]$. Then $LF$ is a splitting field over $F$ of $f(X)$, which is separable over $L$, so $LF/F$ is Galois. To show $\Gal(LF/F)≅\Gal(L/L∩F)$ by restricting the domain from $LF$ to $L$, we essentially repeat the proof of Theorem 2.1a. Consider the restriction homomorphism $$\tag{2.3}\Gal(LF/F) → \Gal(L/K)\text{, where }σ↦σ|L.$$ This has a trivial kernel: an automorphism in $\Gal(LF/F)$ that is trivial on $L$ is trivial on $LF$ since it is automatically trivial on $F$. Therefore (2.3) is injective. The image of (2.3) is a subgroup of $\Gal(L/K)$, so the image has the form $\Gal(L/E)$ for some field $E$ between $K$ and $L$. By Galois theory $E$ is the fixed field of the image of (2.3), so $$E =\{x ∈ L : σ(x) = x\text{ for all }σ ∈\Gal(LF/F)\}.$$ An element of $LF$ is fixed by $\Gal(LF/F)$ precisely when it belongs to $F$, so E = $L∩F$. Therefore the image of (2.3) is $\Gal(L/L∩F)$.

Convolution and the Wiener algebra revisited (a) Let $f, g ∈ \mathrm{L}^2(ℝ)$. Show that \[ (f * g)(x) ≝ ∫_{ℝ} f(x-y) g(y) \mathrm{d} y \] is well-defined for all $x ∈ ℝ$ and that the definition is independent of the chosen representatives for $f$ and $g$ used to compute the integral. Using that $\mathscr{D}(ℝ)$ is dense in $\mathrm{L}^2(ℝ)$ show that $f * g ∈ \mathrm{C}_0(ℝ)^1$. Next, show that the above formula for $f * g$ is consistent with the one obtained from the extended convolution rule: $f * g=ℱ^{-1}(\widehat{f} \widehat{g})$. Deduce that $f * g ∈ ℱ\left(\mathrm{L}^1(ℝ)\right)$. Solution. Take $f_n,g_n∈𝒟(ℝ)$ such that $f_n→f,g_n→g$ in $\mathrm{L}^2(ℝ)$. Want to show $f_n*g_n→f*g$. we have $f_n*g_n∈C^∞(ℝ)$ and $\operatorname{supp}(f_n*g_n)⊆\operatorname{supp}(f_n)+\operatorname{supp}(g_n)$, so $f_n*g_n∈𝒟(ℝ)$. \[|f_n*g_n(x)-f*g(x)|=\left|\int_{ℝ}f_n(x-y)g_n(y)-f(x-y)g(y)dy\right|\\ =\left|\int_{ℝ}f_n(x-y)[g_n(y)-g(y)]dy\right|+\left|\int_ℝ[f_n(x-y)-f(x-y)]g(y)dy\right|\\ ≤‖f_n‖_{L^2}‖g_n-g‖_{L^2}+‖f_n-f‖_{L^2}‖g‖_{L^2}\\ ≤C(‖g_n-g‖_{L^2}+‖f_n-f‖_{L^2}) →0\text{ as }n→∞\] so $f*g∈C_0(ℝ)$, using $C_0(ℝ)=\overline{𝒟(R)}^{‖⋅‖_\sup}$. $f*g=ℱ^{-1}(\hat{f}\hat{g})$ $f_n*g_n→f*g$ uniformly, $f_n*g_n→f*g$ in $𝒮'(ℝ)$. $ℱ(f_n*g_n)→ℱ(f*g)$ in $𝒮'(ℝ)$ By Plancherel $‖\hat{f_n}-\hat{f}‖_{L^2}=\sqrt{2π}‖f_n-f‖_{L^2}$ we have $\hat{f_n}→\hat{f},\hat{g_n}→\hat{g}$ in $L^2$. \[‖\hat{f}\hat{g}-\hat{f_n}\hat{g_n}‖_{L^1}=‖\hat{f}(\hat{g}-\hat{g_n})+(\hat{f}-\hat{f_n})\hat{g_n}‖_{L^1}\\≤‖\hat{f}(\hat{g}-\hat{g_n})‖_{L^1}+‖(\hat{f}-\hat{f_n})\hat{g_n}‖_{L^1}→0\] so $f*g=ℱ^{-1}(\hat{f}\hat{g})$ (b) Show that $2 π ℱ(h k)=\widehat{h} * \widehat{k}$ holds for $h, k ∈ \mathrm{L}^2(ℝ)$, for instance, by use of the convolution rule from (a) together with the Fourier inversion formula in $\mathscr{S}'$. Prove that $ℱ(\mathrm{L}^1(ℝ))=\{h * k: h, k ∈ \mathrm{L}^2(ℝ)\}$. Solution. $\hat{h}*\hat{k}=ℱ^{-1}(\hat{\hat{h}}\hat{\hat{k}})=ℱ^{-1}((2π)^2hk)=2πℱ(hk)$. $f∈L^1(ℝ),h=\operatorname{sgn}(f)|f|^{1/2},k=|f|^{1/2}$ $h,k∈L^2$ $f=\frac1{2π}h*k$ so $ℱ(\mathrm{L}^1(ℝ))=\{h * k: h, k ∈ \mathrm{L}^2(ℝ)\}$.

(a) Let $X$ be a real Banach space, $Y$ and $Z$ be real normed vector spaces, and $B: X × Y → Z$ be bilinear (i.e. linear in each variable). Suppose that for each $x ∈ X$ and $y ∈ Y$, the linear maps $B^x: Y → Z$ and $B_y: X → Z$ defined by \[ B^x(y)=B(x, y)=B_y(x) \]are continuous. Use the principle of uniform boundedness to prove that there exists a constant $K$ such that $‖B(x, y)‖ ≤ K‖x‖‖y‖$ for all $x ∈ X$ and $y ∈ Y$. Deduce that $B$ is continuous. Proof: Let $S^Y=\{y∈Y:‖y‖<1\}$. For each $x ∈ X$, $‖B_y(x)‖=‖B^x(y)‖≤‖B^x‖‖y‖$, so \[\sup_{y∈S^Y}‖B_y(x)‖≤‖B^x‖<∞\] By the principle of uniform boundedness, \[K≔\sup_{y∈S^Y}‖B_y‖<∞\] So $‖B(x,\frac{y}{‖y‖})‖ ≤ K‖x‖$ for each $x∈X,y∈Y$. So $‖B(x,y)‖ ≤ K‖x‖‖y‖$ for each $x∈X,y∈Y$. So $B$ is continuous. (b) Let $X$ and $Y$ both be the subspace of $L^1(0,1)$ consisting of polynomials, $Z=ℝ$, and \[B(f,g)=∫_0^1 f g\] Show that the bilinear form $B$ is continuous in each variable but it is not continuous. {\color{gray}[To put things in perspective, please note that even on $ℝ^2$, for nonlinear functions, separate continuity does not imply joint continuity. A standard example is the function $f(x, y)=\frac{x y}{x^2+y^2}$ for $(x, y) ≠ 0$ and $f(0,0)=0$.]} Proof: By Hölder’s inequality, $|B(f,g)|≤‖f‖_{L^∞}‖g‖_{L^1}$, so $B$ is continuous in $g$, similarly continuous in $f$. For $f(x)=\sqrt{n}χ_{[0,1/n]}$, $‖f‖_{L^1}=1/\sqrt{n}→0$, yet $B(f,f)=1$, so does not exist $K$, so $B$ is not continuous.

Show that the curve $y^2 z=x^3+x z^2$ in $ℂℙ^2$ is nonsingular. Now consider this curve over the finite field $ℤ_p$, where $p$ is a prime. That is, we consider the curve in $P\left((ℤ_p)^3\right)$ with equation $y^2 z=x^3+x z^2$. For which $p$ is this nonsingular? Solution. $P(x,y,z)=y^2 z-x^3-x z^2,\frac{∂P}{∂x}=-3x^2-z^2,\frac{∂P}{∂y}=2yz,\frac{∂P}{∂z}=y^2-2xz$ At singular points, $\frac{∂P}{∂y}=0$, so $y=0$ or $z=0$. If $y=0$, $\frac{∂P}{∂z}=-2xz$, so $x=0$ or $z=0$, then $\frac{∂P}{∂x}=0$ implies $x=y=z=0$❌ If $z=0$, $\frac{∂P}{∂x}=-3x^2=0$, $\frac{∂P}{∂z}=y^2=0$, then $x=y=z=0$❌ So the curve $y^2 z=x^3+x z^2$ in $ℂℙ^2$ is nonsingular. For $p>3$ the above shows the curve over $ℤ_p$ is nonsingular. For $p=2$, $\frac{∂P}{∂x}=x^2+z^2,\frac{∂P}{∂y}=0,\frac{∂P}{∂z}=y^2$, so $[1:0:1]$ is a singular point. For $p=3$, $\frac{∂P}{∂x}=-z^2,\frac{∂P}{∂y}=2yz,\frac{∂P}{∂z}=y^2-2xz\implies$solution is $[1:0:0]$, but it doesn't lie on the curve, so the curve is nonsingular. [over a finite field $d=0$, $\frac{∂P}{∂x}=\frac{∂P}{∂y}=\frac{∂P}{∂z}=0$ don't imply $P=0$.]

(a) Let $X$ be a topological space with base-point $x_{0}$. Define the elements of its fundamental group $\pi_{1}\left(X, x_{0}\right)$. How does $\pi_{1}\left(X, x_{0}\right)$ depend on the choice of base-point? What is the fundamental group of the circle $S^{1}$? For each element give an explicit representative $\alpha:[0,1] \rightarrow S^{1}$. Given a continuous map of spaces $f: X \rightarrow Y$ define the induced map on fundamental groups. Explain what it means to say that the fundamental group is a homotopy invariant. Answers: Elements are based homotopy classes of paths $\alpha: I \rightarrow X$ that start and end at $x_{0}$. If $x_{0}^{\prime}$ is another base point then the two fundamental groups are isomorphic if $x_{0}$ and $x_{0}^{\prime}$ are in the same path component; otherwise they are unrelated. $\pi_{1}(S^{1})=\mathbb{Z}$; $n$ is a represented by $\alpha(t)=e^{2 n \pi i t}$. $f_{*}([\alpha]):=[f \circ \alpha]$. This assumes that if $x_{0}$ is the basepoint for $X$ then $f(x_{0})$ is the basepoint for $Y$. If $X$ and $Y$ are homotopy equivalent spaces then their fundamental groups are isomorphic. (b) State and prove the Brouwer Fixed Point Theorem for continuous maps $$ h: D^{2} \rightarrow D^{2} $$ Answers: [bookwork] (c) Let $p(z)=z^{n}+a_{1} z^{n-1}+\cdots+a_{n}$ be a non-constant polynomial with coefficients in $\mathbb{C}$ and assume (to get a contradiction) that $p(z)$ has no roots in $\mathbb{C}$. (i) For $r \geqslant 0$ let $$ f_{r}(s)=\frac{p(r \exp 2 \pi i s) / p(r)}{|p(r \exp 2 \pi i s) / p(r)|} $$ Explain carefully how each $f_{r}$ defines a based map $S^{1} \rightarrow S^{1}$ which is homotopic to the constant map. (ii) Show that for $r$ large enough, $$ p_{t}(z)=z^{n}+t(a_{1} z^{n-1}+\cdots+a_{n}) $$ has no roots on the circle $|z|=r$ for $0 \leqslant t \leqslant 1$, and hence $p$ may be replaced by $p_{t}$ in the formula for $f_{r}$. Deduce the Fundamental Theorem of Algebra. Answers: (i) $f_{r}$ is well-defined as $p$ has no roots. $f_{r}(1)=1$ and hence it is a based map. $f_{r}(s)$ is continuous both in $r$ and in $s$. Hence all the $f_{r}$ are homotopic to each other. $f_{0}$ is independent of $s$ and hence a constant map. (ii) For $r=|z|$ large enough and $t$ bounded $$ \left|p_{t}(z)\right|=\left|z^{n}+t\left(a_{1} z^{n-1}+\cdots+a_{n}\right)\right| \geqslant\left|z^{n}\right|-\left|t\left(a_{1} z^{n-1}+\cdots+a_{n}\right)\right|>0 $$ and hence has no roots. Replacing $p_{t}$ for $p$, we see that each $p_{t}$ is homotopic to a constant map. But varying $t$ from 0 to 1 we also see that $p_{t}$ is homotopic to $z^{n}$. The Fundamental Theorem of Algebra now follows as $p=p_{1}$ is homotopic to $z^{n}$ and to a constant map. But $z^{n}$ represents $n$ in $\pi_{1}\left(S^{1}\right)$ while the constant map represents 0. Since $p$ is non-constant, $n \neq 0$ and this leads to a contradiction. Hence, our assumption was wrong and $p$ has a root.

THEOREM 3.1. There are two triangles $\Delta_A, \Delta_B$ in the hyperbolic (or Lobachevsky) plane such that $1^{\circ}$ the length of each side of $\Delta_A$ does not exceed the length of the corresponding side of $\Delta_B$, $2^{\circ} \Delta_B$ is non-obtuse, but \[ \text { area } \Delta_A>\text { area } \Delta_B \text {. } \] Proof. Let $\Delta_A$ be an equilateral triangle with each of its angles equal to $\pi / 6$; and let $\Delta_B$ be a right isosceles triangle, with the two sides adjacent to the right angle of the same length as a side of $\Delta_A$. First of all, as we know, in hyperbolic spaces the hypotenuse of a right triangle is the longest side, so we have $1^{\circ}$ the length of each side of $\Delta_A$ does not exceed the length of the corresponding side of $\Delta_B$. Next, since $\Delta_B$ is a right triangle, and because the sum of the interior angles is less than $\pi$ in any triangle, $\Delta_B$ can not have an obtuse angle, that is to say $2^{\circ} \Delta_B$ is non-obtuse. Finally, compare the areas of $\Delta_A$ and $\Delta_B$. As the defect of $\Delta_A$ is equal to $\pi-3\times\pi/6=\pi / 2$, we have \[ \text { area } \Delta_A=\frac{1}{|K|} \frac{\pi}{2} \text {, } \] where $K$ is the curvature of the hyperbolic plane. On the other hand, the sum of the interior angles of $\Delta_B$ is greater than $\pi / 2$, hence its defect is less than $\pi / 2$, and we have \[ \text { area } \Delta_B<\frac{1}{|K|} \frac{\pi}{2} \text {. } \] Comparing these, we obtain \[ \text { area } \Delta_A>\text { area } \Delta_B \text {, } \] and the proof of Theorem 3.1 is completed.

The Wiener algebra $ℱ(\mathrm{L}^1)$ is strictly smaller than $\mathrm{C}_0$. (a) Prove that for $0Proof. $\frac{\sin ξ}{ξ}>0$ on $[2Nπ,(2N+1)π]$, wlog $s=2nπ,t=(2N+1)π\ (n,N∈ℕ_0)$ \[∫_s^t \frac{\sin ξ}{ξ} \mathrm{d} ξ \\<\sum_{k=n}^N\frac1{π(2k-1)}\int_{(2k-1)π}^{2kπ}\sinξ\ \mathrm{d}ξ+\frac1{π (2k+1)}\int_{2kπ}^{(2k+1)π}\sinξ\ \mathrm{d}ξ \\=\frac2π\sum_{k=n}^N\frac1{2k-1}-\frac1{2k+1} \\<\frac2π\sum_{k=1}^∞\frac1{2k-1}-\frac1{2k+1} \\=\frac2π \] (b) Assume that $f ∈ \mathrm{L}^1(ℝ)$ is odd as a distribution (meaning that $\tilde{f}=-f$ holds). Prove that \[ \left|∫_s^t \frac{\widehat{f}(ξ)}{ξ} \mathrm{d} ξ\right| ≤ 4\|f\|_1 \] holds for all $0Proof. \[ \widehat{f}(ξ)=∫_{-∞}^∞f(x)(e^{-ixξ}-e^{ixξ})\mathrm{d}x\\=∫_0^∞f(x)(e^{-ixξ}-e^{ixξ})\mathrm{d}x\\=-2i\int_{0}^{∞}f(x)\sin(ξx)\mathrm{d}x \] then \[ \left|∫_s^t \frac{\widehat{f}(ξ)}{ξ} \mathrm{d} ξ\right|=\left|∫_s^t\frac{2i}{ξ}\int_0^{∞}f(x)\sin(ξx)\mathrm{d}x\mathrm{d}ξ\right|\\ =2\left|\int_0^{∞}f(x)∫_s^t\frac1{ξ}\sin(ξx)\mathrm{d}ξ\mathrm{d}x\right|\\ ≤2\left|\int_0^{∞}f(x)\mathrm{d}x\right|\left|∫_s^t\frac1{ξ}\sin(ξx)\mathrm{d}ξ\right|\\ \overset{\text{(1)}}≤8\int_{0}^{∞}|f(x)|dx\\ =4\int_{-∞}^{∞}|f(x)|dx\\ =4‖f‖_{L^1} \] (c) Let $g ∈ \mathrm{C}_0(ℝ)$ be an odd function satisfying $g(ξ)=1 / \log (ξ)$ for $ξ ≥ 2$. Prove that there does not exist an integrable function whose Fourier transform is $g$. Proof. Supose $∃f∈L^1$ such that $\hat{f}=g$. $f=ℱ^{-1}g=\frac1{2π}\tilde{ℱ}g$ By (b) $\left|∫_s^t \frac{\widehat{f}(ξ)}{ξ} \mathrm{d} ξ\right| ≤4\|f\|_{L^1}<∞$ for all $0

Let $1≤pProof. since $|x_k|^q<|x_k|^p$ for $k$ large, $\sum_k|x_k|^p<∞\implies\sum_k|x_k|^q<∞$, so $l^p(ℝ)⊆l^q(ℝ)$. Consider $A_n≔\{x ∈ l^p:‖x‖_p ≤ n\}$. Want to show $A_n$ is nowhere dense in $l^q$. Thus $l^p=\bigcup_{n=1}^∞A_n$ is of first category in $l^q$. Wlog $n=1$. To prove $A_1$ is closed, if a sequence $x_n∈A_1$ converges to $x∈l^q$, for any $i∈ℕ$, ${|x^{(i)}-x_n^{(i)}|}≤\|x-x_n\|_q → 0$, so $x_n^{(i)} → x^{(i)}$ as $n → ∞$. By Fatou's lemma [Proof: $∀N,n^p≥‖x_n‖_p^p≥\sum_{k≤N}|x_k^{(j)}|^p→\sum_{k≤N}|x_k|^p$.] \[\|x\|_p≤\lim_{n→∞}\|x_n\|_p≤1\]So $A_1$ is closed. To prove $\mathring{A_1}=∅$ in $l^q$, for any $x ∈ A_1$, for any $r$, we show $B_r(x)⊄l^p$: Set $y_j=\frac1{j^{1/p}},y∈ l^q ∖ l^p$, $‖y‖_q=1$, then $\frac{r}{2}y+x$ is in $B_r(x)$ but not in $l^p$, since $y∉l^p$. As $r>0$ is arbitrary, $x∉\mathring{A_1}$ and thus $\mathring{A_1}=∅$. In general every proper subspace has empty interior (2) $l^p$ is dense in $(l^q(ℝ),‖⋅‖_{l^q})$. Proof: The space $c_{00}=\{(x_n):x_n=0 \text{ all but finitely many }n\}$ is dense in $l^q$ $∀1≤q<∞$. Obviously $c_{00}⊂ l^p$ so $l^p$ contains dense subset and is thus dense in $l^q$. %https://math.stackexchange.com/questions/2157682/

Find explicitly the projective transformation of $ℂℙ^2$ that takes the points $[1,0,0],[1,1,0],[0,0,1]$ and $[0,1,1]$ to $[1,2,3],[1,0,1],[0,1,0]$ and $[1,0,-1]$ respectively. Solution. First we scale the points to satisfy linear relations \begin{gather*}[1,0,0]-[1,1,0]-[0,0,1]+[0,1,1]=0\cr[1,2,3]-2[1,0,1]-2[0,1,0]+[1,0,-1]=0\end{gather*} To find a linear map $T\colonℂ^2→ℂ^2$ \begin{align*} T(1,0,0)&=(1,2,3)\\ T(-1,-1,0)&=(-2,0,-2)\\ T(0,0,-1)&=(0,-2,0) \end{align*} by row reduction \begin{align*} T(1,0,0)&=(1,2,3)\\ T(0,1,0)&=(1,-2,-1)\\ T(0,0,1)&=(0,2,0) \end{align*} so the projective transformation $\tilde{T}$ induced by the linear map \[T(x,y,z)=(x+y,2x-2y+2z,3x-y)\] satisfies required conditions. Remark. For $T:V_1→V_2$, dual map $T^*:V_2^*→V_1^*$