To Show that a Certain Function is Smooth

In summary, a smooth function is one that is continuous and has derivatives of all orders at every point in its domain. To show that a function is smooth, it must be continuous and have derivatives of all orders at every point in its domain. The necessary conditions for a function to be smooth are that it must be continuous, have derivatives of all orders at every point, and have a well-defined and continuous slope at every point. A function can be smooth at some points and not at others, and it is important to show that a function is smooth as it indicates a well-behaved and accurately modeled function in various fields.
  • #1
caffeinemachine
Gold Member
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Let $f:\mathbf R^n\times \mathbf R^k\to \mathbf R^n\times \mathbf R^k$ be a diffeomorphism such that:
1) $f(\{\mathbf p\}\times \mathbf R^k)=\{\mathbf p\}\times \mathbf R^k$, for all $\mathbf p\in \mathbf R^n$.
2) $f|\{\mathbf p\}\times\mathbf R^k:\{\mathbf p\}\times \mathbf R\to \{\mathbf p\}\times \mathbf R^k$ is a vector space isomorphism.
To show that there exists $\tau:\mathbf R^n\to GL_k(\mathbf R)$ such that $f(\mathbf p,\mathbf v)=(\mathbf p,\tau_{\mathbf p}(\mathbf v))$ for all $(\mathbf p,\mathbf v)\in \mathbf R^n\times \mathbf R^k$.
Further, $\tau$ is smooth.The exsitence of $\tau$ is immediate. What is not trivial here is the smoothness of $\tau$. Here is my attempt.

Fix the standard basis for $\mathbf R^k$.
For each $\mathbf p\in \mathbf R^n$, let $\tau_{\mathbf p}=[a_{ij}^{\mathbf p}]$ with respect to the standard basis.
Then we will be done if we can show that $\mathbf p\mapsto a_{ij}^{\mathbf p}$ is a smooth function, for each $1\leq i,j\leq k$.
We know by hypothesis that the function which sends $(\mathbf p,\mathbf v)$ to $(\sum_{j=1}^ka_{1j}^{\mathbf p}v_j,\ldots,\sum_{j=1}^ka_{kj}^{\mathbf p}v_j)$ is a smooth function.
Thus, for a fixed $i$, $(\mathbf p,\mathbf v)\mapsto \sum_{j=1}^ka_{ij}^{\mathbf p}v_j$ is smooth.
Fix $j$. Setting $v_1=\cdots=v_{j-1}=v_{j+1}=\cdots=v_k=0$, and $v_j=1$, we see that
$\mathbf p\mapsto a_{ij}^{\mathbf p}$ is smooth and hence each $a_{ij}^{\mathbf p}$ is smooth.

Is the above reasoning correct?
Also, can somebody give a coordinate free approach?
 
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  • #2


Yes, your reasoning is correct. Here is a coordinate-free approach:

Let $p\in \mathbf R^n$ be fixed. Since $f|_{\{p\}\times \mathbf R^k}$ is a vector space isomorphism, it must send linearly independent vectors to linearly independent vectors.
Thus, for any linearly independent set $\{\mathbf v_1,\ldots,\mathbf v_k\}\subset \mathbf R^k$, we have that $\{f(p,\mathbf v_1),\ldots,f(p,\mathbf v_k)\}$ is also a linearly independent set.
Since $f(p,\mathbf v_i)\in \{p\}\times \mathbf R^k$ for all $i$, we can write $f(p,\mathbf v_i)=(p,\tau_i(p))$ for some $\tau_i(p)\in \mathbf R^k$.
Now, define $\tau:\mathbf R^n\to GL_k(\mathbf R)$ by $\tau(p)=[\tau_1(p),\ldots,\tau_k(p)]$.
Then, for any $(p,\mathbf v)\in \mathbf R^n\times \mathbf R^k$, we have $f(p,\mathbf v)=(p,\tau(p)\mathbf v)$.
Since $\tau(p)$ is just a matrix, it is clear that $\tau$ is smooth.
 

FAQ: To Show that a Certain Function is Smooth

What does it mean for a function to be smooth?

A smooth function is one that is continuous and has derivatives of all orders at every point in its domain. This means that the function is well-behaved and has no abrupt changes or breaks in its graph.

How can I show that a function is smooth?

To show that a function is smooth, you must demonstrate that it is continuous and has derivatives of all orders at every point in its domain. This can be done through various methods such as using the definition of continuity and differentiability, or by showing that the function satisfies the necessary conditions for smoothness.

What are the necessary conditions for a function to be smooth?

The necessary conditions for a function to be smooth are that it must be continuous and have derivatives of all orders at every point in its domain. Additionally, the function must also have a well-defined and continuous slope at every point, which ensures that it is not only continuous but also differentiable.

Can a function be smooth at some points but not at others?

Yes, a function can be smooth at some points and not at others. For a function to be considered smooth, it must be continuous and have derivatives of all orders at every point in its domain. This means that the function can have points where it is not smooth, such as points of discontinuity or points where the derivatives do not exist.

Why is it important to show that a function is smooth?

Showing that a function is smooth is important because it indicates that the function is well-behaved and has a well-defined behavior at every point in its domain. This is useful in many fields, such as mathematics, physics, and engineering, as it allows for more accurate modeling and analysis of the function's behavior and properties.

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