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: \[ (f^{-1})'(y) = \left[f'\bigl(f^{-1}(y)\bigr)\right]^{-1} \] ∎
English explanation and missing steps
Take \(y=f(x)\in f(B)\), where \(B\subset A\) is open. The inverse function theorem gives an open neighborhood \(U\subset B\) of \(x\) on which \(f\) is a diffeomorphism onto an open \(V\ni y\). Since \(V=f(U)\subset f(B)\), every point of \(f(B)\) is interior; hence \(f(B)\) is open. Set \(B=A\) for \(f(A)\).
Injectivity on all of \(A\) makes the global inverse well-defined. Near each \(y\), it agrees with the differentiable local inverse, because the preimage is unique. Thus \(f^{-1}\) is differentiable everywhere on \(f(A)\), with derivative \([Df(f^{-1}(y))]^{-1}\). The crucial bridge is agreement of local and global inverses.
Hinglish mein poora explanation
\(y=f(x)\in f(B)\) lo, jahan \(B\subset A\) open hai. IFT se \(x\) ka chhota \(U\subset B\) milta hai jiska image open \(V\ni y\) hai. \(V\subset f(B)\), to \(f(B)\) ke har point ke around open neighborhood hai: image open. \(B=A\) se \(f(A)\) open.
Global injectivity se inverse unique hai. Har \(y\) ke paas IFT ka local smooth inverse milta hai; uniqueness ki wajah se wahi global inverse ka local restriction hai. Isliye global inverse differentiable hai aur derivative \([Df(f^{-1}(y))]^{-1}\).
(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
English explanation and missing steps
(a) A nonconstant injective \(C^1\) function \(\mathbb R^2\to\mathbb R\) must have some nonzero partial derivative. Add the unused coordinate: \(G(x,y)=(f(x,y),y)\). Its local image is open by IFT, so it contains a short vertical segment. Two points on that segment have different preimages but the same first coordinate \(f\), contradicting injectivity.
(b) Choose the maximum rank \(k\le m
Hinglish mein poora explanation
(a) Injective scalar \(C^1\) map nonconstant hai, to kahin partial nonzero. \(G(x,y)=(f(x,y),y)\) banao. IFT se local image open hai; usmein chhota vertical segment hai. Segment ke do points ke preimages alag, par first coordinate \(f\) same: contradiction.
(b) Maximum Jacobian rank \(k\le 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. ∎
English explanation and missing steps
In one dimension, equal values at \(x_1
Hinglish mein poora explanation
1D mein agar \(f(x_1)=f(x_2)\), Rolle theorem beech mein \(f'=0\) dega. Isliye everywhere nonzero derivative se injectivity. 2D example mein Jacobian determinant \(e^{2x}(\cos^2y+\sin^2y)=e^{2x}>0\). Phir bhi \((x,y)\) aur \((x,y+2\pi)\) ka image same hai. Local invertibility angular periodicity ko remove nahi karti.
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. ∎
English explanation and missing steps
At zero, the difference quotient is \(1/2+h\sin(1/h)\to1/2\). For \(x\ne0\), \(f'(x)=1/2+2x\sin(1/x)-\cos(1/x)\). Along \(x_n=1/(2n\pi)\), derivatives equal \(-1/2\); along \(y_n=1/((2n+1)\pi)\), they equal \(3/2\). So \(f'\) is discontinuous at zero.
Why no local inverse? Every neighborhood of zero contains points where the derivative is positive and points where it is negative. A continuous one-to-one function on an interval must be strictly monotone; a differentiable strictly increasing function cannot have a negative derivative anywhere, and a strictly decreasing one cannot have a positive derivative. Thus this \(f\) is not one-to-one on any neighborhood of zero, even though \(f'(0)\ne0\). This completes the counterexample beyond merely observing discontinuity of \(f'\).
Hinglish mein poora explanation
Origin par quotient \(1/2+h\sin(1/h)\to1/2\), isliye \(f'(0)\ne0\). Bahar \(f'(x)=1/2+2x\sin(1/x)-\cos(1/x)\). Sequence \(1/(2n\pi)\) par derivative \(-1/2\), aur \(1/((2n+1)\pi)\) par \(3/2\); derivative continuous nahi.
Missing final step: Origin ke har neighborhood mein positive aur negative derivative dono milte hain. Interval par continuous injective function strictly monotone hota hai. Increasing differentiable function ka derivative negative nahi ho sakta, decreasing ka positive nahi. Isliye yeh function kisi bhi origin-neighborhood par one-to-one nahi, despite \(f'(0)\ne0\). Sirf derivative discontinuity bolna proof complete nahi karta.