Help with D&K Lemma 2.2.7 (Hadamard) Proof: Qs on Continuity of \phi_a

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 2: Differentiation ... ...

I need help with another aspect of the proof of Lemma 2.2.7 (Hadamard...) ... ...

Duistermaat and Kolk's Lemma 2.2.7 and its proof read as follows:https://www.physicsforums.com/attachments/7837
https://www.physicsforums.com/attachments/7838
Near to the end of the above text D&K write the following:

" ... ... A direct computation gives \(\displaystyle \| \epsilon_a(h) h^t \|_{ Eucl } = \| \epsilon_a(h) \| \| h \|\), hence\(\displaystyle \lim_{ h \rightarrow 0 } \frac{ \| \epsilon_a(h) h^t \|_{ Eucl } }{ \| h \|^2 } = \lim_{ h \rightarrow 0 } \frac{ \| \epsilon_a(h) \| }{ \| h \| } = 0 \)This shows that \(\displaystyle \phi_a\) is continuous at \(\displaystyle a\). ... ... "

My questions are as follows:

Question 1

... how/why does the above show that \(\displaystyle \phi_a\) is continuous at \(\displaystyle a\). ... ...?

Can someone please demonstrate explicitly, formally and rigorously that \(\displaystyle \phi_a\) is continuous at \(\displaystyle a\). ... ...?Question 2

How/why does the proof of Hadamard's Lemma 2.2.7 imply that \(\displaystyle f\) is continuous at \(\displaystyle a\) if \(\displaystyle f\) is differentiable at \(\displaystyle a\) ... ?
Help will be much appreciated ... ...

Peter==========================================================================================

NOTE:

The start of D&K's section on differentiable mappings may help readers of the above post understand the context and notation of the post ... so I am providing the same as follows:
View attachment 7839
View attachment 7840
The start of D&K's section on linear mappings may also help readers of the above post understand the context and notation of the post ... so I am providing the same as follows:

View attachment 7841
View attachment 7842
View attachment 7843
Hope the above helps readers understand the context and notation of the post ...

Peter

 

[GJA]

Hi, Peter.

Question 1

... how/why does the above show that \(\displaystyle \phi_a\) is continuous at \(\displaystyle a\). ... ...?

Can someone please demonstrate explicitly, formally and rigorously that \(\displaystyle \phi_a\) is continuous at \(\displaystyle a\). ... ...?

This question is perhaps a little deeper than it seems initially. The subtlety lies in the fact that $\phi_{a}(x)$ is an operator-valued function; i.e., for each $x$, $\phi_{a}(x)$ is a linear operator.

Step 1: Understanding Continuity
When the authors say "$\phi_{a}$ is continuous at $a$," what they mean is the operators $\phi_{a}(x)$ and $\phi_{a}(a)$ can be made arbitrarily close for $x$ sufficiently close to $a$ (i.e., the "distance" from $\phi_{a}(x)$ to $\phi_{a}(a)$ vanishes in the limit as $x\rightarrow a$). Symbolically what we want to show is that
$$\lim_{x\rightarrow a}\phi_{a}(x)=\phi_{a}(a)$$
Note that, at least symbolically, this is exactly the same notion as continuity in the case of a real-valued function of a single variable; i.e., $f$ is continuous at $a$ means
$$\lim_{x\rightarrow a}f(x)=f(a).$$
Step 2: Understanding Operator Norms
This begs the question: How do we measure the distance between operators? Depending on the context, there are several ways one can define a norm on a space of functions (this is not unlike your previous posts where you asked about different norms on $\mathbb{R}^{n}$). In the case of your post, the authors are (implicitly) using a notion known as "strong (operator) continuity," where, by definition,
$$\lim_{x\rightarrow a}\phi_{a}(x)=\phi_{a}(a)$$
provided
$$\lim_{x\rightarrow a}\|\phi_{a}(x)y-\phi_{a}(a)y\|=0,\qquad\forall y\in\mathbb{R}^{n},$$
where the norm above is the norm in $\mathbb{R}^{p}$ (which makes sense because $\phi_{a}(x)$ and $\phi_{a}(a)$ are operators taking elements of $\mathbb{R}^{n}$ into $\mathbb{R}^{p}$). Note that this condition which defines the meaning of the limit is the reason the authors need to introduce a $y$ in the equation
$$\phi_{a}(x)y=Df(a)y+\cdots$$
Step 3: Answering Your Question
From what was discussed in Step 2, we must show
$$\lim_{x\rightarrow a}\|\phi_{a}(x)y-\phi_{a}(a)y\|=0,\qquad\forall y\in\mathbb{R}^{n}.$$
Note that, by the authors' definition of $\phi_{a}(x)$, $\phi_{a}(a)=Df(a)$. Looking at the equation
$$\phi_{a}(x)y=Df(a)y+\cdots,$$
we see $\phi_{a}(x)-\phi_{a}(a)=\phi_{a}(x)-Df(a)=\cdots,$ so when the authors show
$$\lim_{h\rightarrow 0}\frac{\|\epsilon_{a}(h)h^{t}\|}{\|h\|^{2}}=0,$$
they are really proving
$$\lim_{x\rightarrow a}\|\phi_{a}(x)y-\phi_{a}(a)y\|=0,\qquad\forall y\in\mathbb{R}^{n},$$
which is precisely what we want to establish when proving $\phi_{a}(x)$ is continuous at $a$.

Question 2
How/why does the proof of Hadamard's Lemma 2.2.7 imply that \(\displaystyle f\) is continuous at \(\displaystyle a\) if \(\displaystyle f\) is differentiable at \(\displaystyle a\) ... ?

To prove that $f$ is continuous at $a$, we want to show that
$$\lim_{x\rightarrow a}f(x)=f(a),$$
where we note that $f$ is a function, not an operator, so we don't need to introduce $y$ as we did previously. From the above we know
$$\lim_{x\rightarrow a}\phi_{a}(x)=\phi_{a}(a)=Df(a),$$
so
$$\lim_{x\rightarrow a} f(x)=\lim_{x\rightarrow a} f(a)+\lim_{x\rightarrow a}\phi_{a}(x)(x-a)=f(a)+Df(a)(a-a)=f(a),$$
as desired.

 

[Math Amateur]

Hi, Peter.
This question is perhaps a little deeper than it seems initially. The subtlety lies in the fact that $\phi_{a}(x)$ is an operator-valued function; i.e., for each $x$, $\phi_{a}(x)$ is a linear operator.

Step 1: Understanding Continuity
When the authors say "$\phi_{a}$ is continuous at $a$," what they mean is the operators $\phi_{a}(x)$ and $\phi_{a}(a)$ can be made arbitrarily close for $x$ sufficiently close to $a$ (i.e., the "distance" from $\phi_{a}(x)$ to $\phi_{a}(a)$ vanishes in the limit as $x\rightarrow a$). Symbolically what we want to show is that
$$\lim_{x\rightarrow a}\phi_{a}(x)=\phi_{a}(a)$$
Note that, at least symbolically, this is exactly the same notion as continuity in the case of a real-valued function of a single variable; i.e., $f$ is continuous at $a$ means
$$\lim_{x\rightarrow a}f(x)=f(a).$$
Step 2: Understanding Operator Norms
This begs the question: How do we measure the distance between operators? Depending on the context, there are several ways one can define a norm on a space of functions (this is not unlike your previous posts where you asked about different norms on $\mathbb{R}^{n}$). In the case of your post, the authors are (implicitly) using a notion known as "strong (operator) continuity," where, by definition,
$$\lim_{x\rightarrow a}\phi_{a}(x)=\phi_{a}(a)$$
provided
$$\lim_{x\rightarrow a}\|\phi_{a}(x)y-\phi_{a}(a)y\|=0,\qquad\forall y\in\mathbb{R}^{n},$$
where the norm above is the norm in $\mathbb{R}^{p}$ (which makes sense because $\phi_{a}(x)$ and $\phi_{a}(a)$ are operators taking elements of $\mathbb{R}^{n}$ into $\mathbb{R}^{p}$). Note that this condition which defines the meaning of the limit is the reason the authors need to introduce a $y$ in the equation
$$\phi_{a}(x)y=Df(a)y+\cdots$$
Step 3: Answering Your Question
From what was discussed in Step 2, we must show
$$\lim_{x\rightarrow a}\|\phi_{a}(x)y-\phi_{a}(a)y\|=0,\qquad\forall y\in\mathbb{R}^{n}.$$
Note that, by the authors' definition of $\phi_{a}(x)$, $\phi_{a}(a)=Df(a)$. Looking at the equation
$$\phi_{a}(x)y=Df(a)y+\cdots,$$
we see $\phi_{a}(x)-\phi_{a}(a)=\phi_{a}(x)-Df(a)=\cdots,$ so when the authors show
$$\lim_{h\rightarrow 0}\frac{\|\epsilon_{a}(h)h^{t}\|}{\|h\|^{2}}=0,$$
they are really proving
$$\lim_{x\rightarrow a}\|\phi_{a}(x)y-\phi_{a}(a)y\|=0,\qquad\forall y\in\mathbb{R}^{n},$$
which is precisely what we want to establish when proving $\phi_{a}(x)$ is continuous at $a$.
To prove that $f$ is continuous at $a$, we want to show that
$$\lim_{x\rightarrow a}f(x)=f(a),$$
where we note that $f$ is a function, not an operator, so we don't need to introduce $y$ as we did previously. From the above we know
$$\lim_{x\rightarrow a}\phi_{a}(x)=\phi_{a}(a)=Df(a),$$
so
$$\lim_{x\rightarrow a} f(x)=\lim_{x\rightarrow a} f(a)+\lim_{x\rightarrow a}\phi_{a}(x)(x-a)=f(a)+Df(a)(a-a)=f(a),$$
as desired.

Thanks GJA ... that was really helpful ...

Still reflecting on what you have said ...

Thanks again ...

Peter