Roll No: 240423
Course: MTH403 Real Analysis II
Assignment: HW-4
Let \(A\subset \mathbf{R}^n\) be an open set. Let \(f:A\to\mathbf{R}^n\) be a continuously differentiable one-to-one function. Assume that \(\det f'(x)\neq 0\) for all \(x\in A\).
Show that:
- \(f(A)\) is an open set.
- \(f^{-1}:f(A)\to A\) is differentiable.
- \(f(B)\) is open for any open set \(B\subset A\).
Let \(B\subset A\) be an open set. We shall prove that \(f(B)\) is open.
Take any \(y\in f(B)\). Then there exists \(x\in B\) such that \(f(x)=y\).
Since \(B\) is open, there exists an open neighborhood \(W\) of \(x\) such that \(W\subset B\).
Also, by hypothesis, \(\det f'(x)\neq0\).
Therefore, by the Inverse Function Theorem, there exist open neighborhoods \(U\) of \(x\) and \(V\) of \(y=f(x)\) such that \(f|_U:U\to V\) is a diffeomorphism. In particular, \(f(U)=V\).
We may choose \(U\subset B\). Hence \(V=f(U)\subset f(B)\).
Thus, for every \(y\in f(B)\), there exists an open neighborhood \(V\) of \(y\) such that \(V\subset f(B)\).
Therefore, \(f(B)\) is open.
Taking \(B=A\), we obtain \(f(A)\) is open. ∎
Since \(f\) is one-to-one, the inverse function \(f^{-1}:f(A)\to A\) exists.
Let \(y\in f(A)\) and put \(x=f^{-1}(y)\). Thus, \(f(x)=y\).
Since \(\det f'(x)\neq0\), the Inverse Function Theorem gives open neighborhoods \(U\) of \(x\) and \(V\) of \(y\), together with a differentiable local inverse \(g:V\to U\) such that \(g=f^{-1}\) on \(V\).
Indeed, for every \(z\in V\), \(f(g(z))=z\). Since \(f\) is one-to-one, the point \(g(z)\) must be the unique point \(f^{-1}(z)\). Hence \(g(z)=f^{-1}(z)\).
Therefore, \(f^{-1}\) is differentiable in a neighborhood of every \(y\in f(A)\).
Consequently, \(f^{-1}:f(A)\to A\) is differentiable.
Moreover, its derivative is given by: \[ (f^{-1})'(y) = \left[f'\bigl(f^{-1}(y)\bigr)\right]^{-1} \] ∎
(a) Let \(f:\mathbf{R}^2\to\mathbf{R}\) be a continuously differentiable function. Show that \(f\) is not one-to-one.
(b) Generalize this to a continuously differentiable function \(f:\mathbf{R}^n\to\mathbf{R}^m\) where \(m
Suppose for contradiction, that \(f\) is one-to-one.
We first observe that \(f\) cannot satisfy \(D_1f(x,y)=0\) at every point. Indeed, if this were the case, then \(f\) would be independent of \(x\), so \(f(x,y)=f(0,y)\), which would contradict the one-to-one property.
Therefore, there exists a point \((a,b)\) such that \(D_1f(a,b)\neq0\).
Since \(D_1f\) is continuous, there exists an open neighborhood \(A\) of \((a,b)\) such that \(D_1f(x,y)\neq0\) for all \((x,y)\in A\).
Define \(g:A\to\mathbf{R}^2\) by \(g(x,y)=\bigl(f(x,y),y\bigr)\).
Its derivative is: \[ g'(x,y) = \begin{pmatrix} D_1f(x,y) & D_2f(x,y)\\ 0 & 1 \end{pmatrix} \]
Therefore, \(\det g'(x,y) = D_1f(x,y)\neq0\).
By the Inverse Function Theorem, \(g(A)\) is open.
Choose \((u,v)\in g(A)\). Since \(g(A)\) is open, there exists \(\varepsilon>0\) such that \((u,v+\varepsilon)\in g(A)\).
Hence, there exist \((x_1,y_1),(x_2,y_2)\in A\) such that \(g(x_1,y_1)=(u,v)\) and \(g(x_2,y_2)=(u,v+\varepsilon)\).
From the definition of \(g\), we have \(f(x_1,y_1)=u\), \(y_1=v\), and \(f(x_2,y_2)=u\), \(y_2=v+\varepsilon\).
Thus, \(f(x_1,y_1)=f(x_2,y_2)=u\). But \(y_1\neq y_2\), so \((x_1,y_1)\neq(x_2,y_2)\).
This contradicts the assumption that \(f\) is one-to-one.
Therefore, \(f:\mathbf{R}^2\to\mathbf{R}\) cannot be one-to-one. ∎
We prove that there is no continuously differentiable one-to-one map \(f:\mathbf{R}^n\to\mathbf{R}^m\) when \(m Suppose, for contradiction, that \(f\) is one-to-one. Let \(k=\max_{x\in\mathbf{R}^n}\operatorname{rank}f'(x)\). Since \(f'(x)\) is an \(m\times n\) matrix, \(k\leq m Choose \(a\in\mathbf{R}^n\) such that \(\operatorname{rank}f'(a)=k\). After a suitable rearrangement of the coordinates in the domain and codomain, write \(x=(u,v)\), where \(u\in\mathbf{R}^k\), \(v\in\mathbf{R}^{n-k}\), and write \(f=(f_1,f_2)\), where \(f_1:\mathbf{R}^n\to\mathbf{R}^k\). Since \(\operatorname{rank}f'(a)=k\), there is a \(k\times k\) minor of \(f'(a)\) which is nonzero. Thus, after choosing the coordinates appropriately, \(\det\left(\frac{\partial f_1}{\partial u}(a)\right)\neq0\). Define \(g(u,v)=\bigl(f_1(u,v),v\bigr)\). Its derivative has the block form:
\[ g'(u,v) = \begin{pmatrix} \dfrac{\partial f_1}{\partial u} & \dfrac{\partial f_1}{\partial v} \\[1em] 0 & I_{n-k} \end{pmatrix} \] Therefore, \(\det g'(a) = \det\left(\frac{\partial f_1}{\partial u}(a)\right) \neq0\). By the Inverse Function Theorem, \(g\) has a differentiable local inverse near \(a\). Let \(h=f\circ g^{-1}\). Since \(f\) is one-to-one and \(g^{-1}\) is one-to-one, \(h\) is one-to-one. The map \(h\) has the form \(h(y,z) = \bigl(y,\widetilde f_2(y,z)\bigr)\), where \(y\in\mathbf{R}^k\), \(z\in\mathbf{R}^{n-k}\). By the maximality of \(k\), \(\operatorname{rank}h'\leq k\). But the first \(k\) components of \(h\) are the coordinates \(y\). Therefore, if \(\frac{\partial\widetilde f_2}{\partial z}\) were nonzero in any direction, the rank of \(h'\) would exceed \(k\), which is impossible. Hence \(\frac{\partial\widetilde f_2}{\partial z}=0\). Thus \(\widetilde f_2\) is independent of \(z\). Hence there exists a function \(\phi\) such that \(h(y,z)=\bigl(y,\phi(y)\bigr)\). Therefore, for fixed \(y\) and any \(z_1\neq z_2\), \(h(y,z_1)=h(y,z_2)\). This contradicts the fact that \(h\) is one-to-one. Therefore, There is no continuously differentiable one-to-one map \(\mathbf{R}^n\to\mathbf{R}^m\) when \(m
(a) Let \(f:\mathbf{R}\to\mathbf{R}\). If \(f'(a)\neq0\) for all \(a\), show that \(f\) is one-to-one.
(b) Define \(f(x,y)=\bigl(e^x\cos y,e^x\sin y\bigr)\). Show that \(\det f'(x,y)\neq0\) everywhere, but \(f\) is not one-to-one.
Suppose to the contrary, that \(f\) is not one-to-one.
Then there exist \(x_1\neq x_2\) such that \(f(x_1)=f(x_2)\).
Without loss of generality, assume \(x_1 Since \(f\) is differentiable, it is continuous. Therefore, Rolle's Theorem applies to \(f\) on \([x_1,x_2]\). Hence there exists \(c\in(x_1,x_2)\) such that \(f'(c)=0\). This contradicts the hypothesis \(f'(a)\neq0\) for every \(a\in\mathbf{R}\). Therefore, \(f\) is one-to-one. ∎
Let \(f(x,y)=\bigl(e^x\cos y,e^x\sin y\bigr)\).
The derivative matrix is: \[ f'(x,y) = \begin{pmatrix} e^x\cos y & -e^x\sin y\\ e^x\sin y & e^x\cos y \end{pmatrix} \]
Therefore, \begin{aligned} \det f'(x,y) &= (e^x\cos y)(e^x\cos y) - (-e^x\sin y)(e^x\sin y) \\ &= e^{2x}\cos^2y + e^{2x}\sin^2y \\ &= e^{2x}\bigl(\cos^2y+\sin^2y\bigr) \\ &= e^{2x}. \end{aligned}
Since \(e^{2x}>0\) for every \(x\in\mathbf{R}\), we have \(\det f'(x,y)\neq0\) for all \((x,y)\in\mathbf{R}^2\).
However, \(f\) is not one-to-one.
Indeed, \(f(0,0) = \bigl(e^0\cos0,e^0\sin0\bigr) = (1,0)\), whereas \(f(0,2\pi) = \bigl(e^0\cos2\pi,e^0\sin2\pi\bigr) = (1,0)\).
But \((0,0)\neq(0,2\pi)\).
Thus two distinct points have the same image. Hence \(f\) is not one-to-one. ∎
Let \[ f(x)= \begin{cases} \dfrac{x}{2}+x^2\sin\left(\dfrac1x\right),&x\neq0,\\[0.5em] 0,&x=0. \end{cases} \] Show why the Inverse Function Theorem needs the derivative to be continuous.
We first calculate the derivative at \(x=0\).
By the definition of the derivative, \[ f'(0) = \lim_{h\to0} \frac{f(h)-f(0)}{h} = \lim_{h\to0} \left( \frac12+h\sin\frac1h \right). \]
Since \(\left|h\sin\frac1h\right| \leq |h|\to0\), we obtain \(f'(0)=\frac12\neq0.\)
Now let \(x\neq0\). Differentiating, \[ f'(x) = \frac12 + 2x\sin\left(\frac1x\right) + x^2\cos\left(\frac1x\right) \left(-\frac1{x^2}\right) = \frac12 + 2x\sin\left(\frac1x\right) - \cos\left(\frac1x\right). \]
Thus, \(f'(x) = \frac12 + 2x\sin\left(\frac1x\right) - \cos\left(\frac1x\right), \qquad x\neq0.\)
Showing that \(f'\) is not continuous at \(0\)
Consider the sequence \(x_n=\frac1{2n\pi}\).
Then \(\cos\left(\frac1{x_n}\right) = \cos(2n\pi)=1\) and \(\sin\left(\frac1{x_n}\right) = \sin(2n\pi)=0\).
Therefore, \(f'(x_n) = \frac12-1 = -\frac12\).
Now consider \(y_n=\frac1{(2n+1)\pi}\).
Then \(\cos\left(\frac1{y_n}\right) = \cos((2n+1)\pi)=-1\) and \(\sin\left(\frac1{y_n}\right) = \sin((2n+1)\pi)=0\).
Hence, \(f'(y_n) = \frac12-(-1) = \frac32\).
Since \(x_n\to0\) and \(y_n\to0\), but \(f'(x_n)=-\frac12\) and \(f'(y_n)=\frac32\), the limit \(\lim_{x\to0}f'(x)\) does not exist.
Therefore, \(f'\) is not continuous at \(0\).
Conclusion
Now, We have \(f'(0)=\frac12\neq0\), but \(f'\) is not continuous at \(0\).
Thus, merely knowing that \(f'(0)\neq0\) does not provide the continuity condition required in the standard Inverse Function Theorem.
The continuity of \(f'\) guarantees that when \(f'(0)\) is invertible, the derivative remains invertible in a sufficiently small neighborhood of \(0\). This is the key point used in the standard proof of the theorem.
Hence this example illustrates the importance of the \(C^1\) hypothesis in the standard Inverse Function Theorem.
The continuity of \(f'\) is an essential hypothesis in the standard IFT. ∎