Equivalent Norms In $C^k[a,b]$: A Detailed Proof

Aug 14, 2025 by Rajiv Sharma 49 views

Equivalent Norms in $C^k[a,b]$: A Comprehensive Guide

Hey guys! Today, we're diving deep into the fascinating world of Real Analysis and Functional Analysis to explore the equivalence of two norms defined on the space $C^k[a, b]$ . This is a crucial concept when dealing with derivatives, normed spaces, and even Taylor expansions. So, buckle up and let's get started!

Understanding $C^k[a,b]$ and Norms

Before we jump into the nitty-gritty details, let's make sure we're all on the same page. The space $C^k[a, b]$ represents the set of all functions that are k-times continuously differentiable on the closed interval $[a, b]$ . In simpler terms, these are functions whose derivatives up to order k exist and are continuous on the interval. Think of smooth, well-behaved functions – no sudden jumps or breaks!

Now, what about norms? A norm is essentially a way to measure the "size" or "length" of a function. It provides a way to quantify how "big" a function is in a given space. In our case, we're interested in two specific norms defined on $C^k[a, b]$ .

Defining the Norms: $||f||_*$ and $||f||_{**}$

We're going to explore two norms, denoted as $||f||_*$ and $||f||_{**}$ . These norms provide different perspectives on measuring the "size" of a function f within the space $C^k[a, b]$ . Let's break down each norm individually:

The First Norm: $||f||_*$

The first norm, denoted by $||f||_*$ , is defined as the sum of the supremum norm (or infinity norm) of the function f itself and the supremum norm of its k-th derivative, $f^{(k)}$ . Mathematically, we can write it as:

$||f||_* = ||f||_\infty + ||f^{(k)}||_\infty$

Let's unpack this a bit further. The supremum norm, $||f||_\infty$ , represents the maximum absolute value of the function f on the interval $[a, b]$ . In other words, it's the largest "height" the function reaches within the interval. Similarly, $||f^{(k)}||_\infty$ represents the maximum absolute value of the k-th derivative of f on the same interval. This gives us an idea of the maximum rate of change of the (k-1)-th derivative.

So, $||f||_*$ essentially captures both the maximum magnitude of the function and the maximum magnitude of its highest-order derivative. It gives us a sense of how "large" the function is and how much its k-th derivative varies.

The Second Norm: $||f||_{**}$

The second norm, denoted by $||f||_{**}$ , takes a more comprehensive approach. It's defined as the sum of the supremum norms of all derivatives of f from order 0 up to order k. In mathematical notation, this is expressed as:

$||f||_{**} = \sum_{j=0}^{k} ||f^{(j)}||_\infty$

Here, $f^{(j)}$ represents the j-th derivative of f, with $f^{(0)}$ simply being the function f itself. So, we're summing the maximum absolute values of the function, its first derivative, its second derivative, and so on, all the way up to its k-th derivative.

In essence, $||f||_{**}$ captures the magnitude of the function and all its derivatives up to order k. This norm provides a more detailed picture of the function's behavior, taking into account the variation of all its derivatives.

Why are these norms important?

These norms are essential tools in functional analysis, providing ways to measure the size and behavior of functions. They are particularly useful when dealing with concepts like convergence and completeness in function spaces. Understanding the relationships between these norms, such as their equivalence, helps us to better understand the properties of the function space $C^k[a, b]$ .

The Obvious Inequality: $||f||_* extbf{\leq} ||f||_{**}$

Okay, guys, let's tackle the first part of our adventure: proving that $||f||_* \leq ||f||_{**}$ . This might seem a bit daunting at first, but trust me, it's actually quite straightforward. We'll break it down step by step to make sure everyone's following along.

Breaking Down the Inequality

Remember our definitions? We have:

$||f||_* = ||f||_\infty + ||f^{(k)}||_\infty$

and

$||f||_{**} = \sum_{j=0}^{k} ||f^{(j)}||_\infty = ||f||_\infty + ||f'||_\infty + ... + ||f^{(k)}||_\infty$

So, what we need to show is:

$||f||_\infty + ||f^{(k)}||_\infty \leq ||f||_\infty + ||f'||_\infty + ... + ||f^{(k)}||_\infty$

The Intuitive Leap

Now, let's think about this intuitively. On the left-hand side, we're adding the supremum norm of the function itself and the supremum norm of its k-th derivative. On the right-hand side, we're adding the supremum norms of all the derivatives, from order 0 up to order k.

Key Insight: The sum on the right-hand side includes all the terms present on the left-hand side, and potentially more (the supremum norms of the intermediate derivatives). Since we're dealing with norms, which are always non-negative, adding more terms can only increase the sum.

The Formal Proof (It's Easier Than You Think!)

Let's put this intuition into a formal proof. It's surprisingly simple:

Start with the definition of $||f||_{**}$ :

$||f||_{**} = \sum_{j=0}^{k} ||f^{(j)}||_\infty$
Expand the summation:

$||f||_{**} = ||f||_\infty + ||f'||_\infty + ... + ||f^{(k-1)}||_\infty + ||f^{(k)}||_\infty$
Notice that $||f||_*$ is a subset of the terms in $||f||_{**}$ :

We can rewrite $||f||_{**}$ as:

$||f||_{**} = (||f||_\infty + ||f^{(k)}||_\infty) + (||f'||_\infty + ... + ||f^{(k-1)}||_\infty)$

The first group of terms is exactly $||f||_*$ .
The Remaining Terms are Non-Negative:

Since norms are always non-negative, the second group of terms $(||f'||_\infty + ... + ||f^{(k-1)}||_\infty)$ is greater than or equal to zero.
Therefore:

$||f||_{**} = ||f||_* + (||f'||_\infty + ... + ||f^{(k-1)}||_\infty) \geq ||f||_*$

Conclusion of this inequality

And that's it! We've successfully shown that $||f||_* \leq ||f||_{**}$ . The key takeaway here is that $||f||_{**}$ includes more information about the function's derivatives, making it naturally larger than $||f||_*$ , which only considers the function and its k-th derivative. This simple inequality lays the groundwork for the more interesting part: proving the equivalence of these norms.

Proving the Equivalence: The Other Direction

Alright, guys, now comes the real challenge! We've already established that $||f||_* \leq ||f||_{**}$ , which is a good start. But to prove that these norms are equivalent, we need to show that there exists a constant C such that $||f||_{**} \leq C||f||_*$ . In other words, we need to show that $||f||_{**}$ is bounded above by a multiple of $||f||_*$ .

This is where things get a bit more interesting, and we'll need to pull out some more advanced tools from our mathematical arsenal. Don't worry, though; we'll break it down into manageable steps.

The Road Map: Taylor's Theorem to the Rescue

The key to proving this inequality lies in a powerful result known as Taylor's Theorem. Taylor's Theorem allows us to approximate a function at a point using its derivatives at another point. This is precisely what we need to relate the intermediate derivatives (those between the 0-th and k-th) to the function itself and its k-th derivative.

Here's the plan:

Apply Taylor's Theorem: We'll use Taylor's Theorem to express the derivatives of f at a point in terms of f and its k-th derivative.
Estimate the Intermediate Derivatives: We'll use the Taylor expansion to bound the supremum norms of the intermediate derivatives, $||f^{(j)}||_\infty$ for $1 \leq j \leq k-1$ .
Sum the Inequalities: We'll sum the inequalities we obtain to get a bound for $||f||_{**}$ in terms of $||f||_*$ .

Step 1: Unleashing Taylor's Theorem

Let's start by stating Taylor's Theorem. For a function $f \in C^k[a, b]$ , Taylor's Theorem tells us that for any $x, x_0 \in [a, b]$ , we can write:

$f(x) = \sum_{j=0}^{k-1} \frac{f^{(j)}(x_0)}{j!}(x - x_0)^j + \frac{1}{(k-1)!} \int_{x_0}^{x} (x - t)^{k-1} f^{(k)}(t) dt$

This might look a bit intimidating, but let's break it down. The left-hand side is just the value of the function at x. The right-hand side is a sum of terms involving the derivatives of f at a fixed point $x_0$ , plus an integral term that captures the "remainder" of the approximation.

Step 2: Estimating the First Derivative

We're particularly interested in estimating the first derivative, $f'(x)$ . To do this, we differentiate the Taylor expansion with respect to x:

$f'(x) = \sum_{j=1}^{k-1} \frac{f^{(j)}(x_0)}{(j-1)!}(x - x_0)^{j-1} + \frac{1}{(k-2)!} \int_{x_0}^{x} (x - t)^{k-2} f^{(k)}(t) dt$

Now, let's take the absolute value of both sides and use the triangle inequality:

$|f'(x)| \leq \sum_{j=1}^{k-1} \frac{|f^{(j)}(x_0)|}{(j-1)!}|x - x_0|^{j-1} + \frac{1}{(k-2)!} \left| \int_{x_0}^{x} (x - t)^{k-2} f^{(k)}(t) dt \right|$

We can further bound the integral term using the fact that $|f^{(k)}(t)| \leq ||f^{(k)}||_\infty$ :

$\left| \int_{x_0}^{x} (x - t)^{k-2} f^{(k)}(t) dt \right| \leq \int_{x_0}^{x} |x - t|^{k-2} |f^{(k)}(t)| dt \leq ||f^{(k)}||_\infty \int_{x_0}^{x} |x - t|^{k-2} dt$

The integral on the right can be easily evaluated, and we obtain a bound for $|f'(x)|$ in terms of the values of $f^{(j)}(x_0)$ and $||f^{(k)}||_\infty$ .

The Trick: Clever Choice of $x_0$

Here's a clever trick: we can choose a specific point $x_0 \in [a, b]$ where we can bound the values of $f^{(j)}(x_0)$ in terms of $||f||_\infty$ . For example, we can use the Mean Value Theorem to find points where $f^{(j)}(x_0)$ is related to the average value of $f^{(j)}$ on the interval, which can then be bounded by $||f||_\infty$ .

By carefully choosing $x_0$ and applying the Mean Value Theorem repeatedly, we can show that there exists a constant M such that:

$|f^{(j)}(x_0)| \leq M||f||_\infty$

for all $0 \leq j \leq k-1$ .

Back to the Estimate for $f'(x)$

Plugging this bound back into our estimate for $|f'(x)|$ , we get:

$|f'(x)| \leq C_1 ||f||_\infty + C_2 ||f^{(k)}||_\infty$

where $C_1$ and $C_2$ are constants that depend on k and the length of the interval $[a, b]$ . Taking the supremum over all x in $[a, b]$ , we obtain:

$||f'||_\infty \leq C_1 ||f||_\infty + C_2 ||f^{(k)}||_\infty \leq \max\{C_1, C_2\} (||f||_\infty + ||f^{(k)}||_\infty) = \max\{C_1, C_2\} ||f||_*$

Step 3: Estimating Higher-Order Derivatives

We can repeat a similar process for the higher-order derivatives $f''(x), f'''(x), ..., f^{(k-1)}(x)$ . By repeatedly differentiating the Taylor expansion and using similar bounding techniques, we can show that for each $1 \leq j \leq k-1$ , there exists a constant $C_j$ such that:

$||f^{(j)}||_\infty \leq C_j ||f||_*$

Step 4: Summing the Inequalities and Reaching the Conclusion

Finally, we can sum all these inequalities to obtain a bound for $||f||_{**}$ :

$||f||_{**} = \sum_{j=0}^{k} ||f^{(j)}||_\infty = ||f||_\infty + ||f^{(k)}||_\infty + \sum_{j=1}^{k-1} ||f^{(j)}||_\infty$

Using our previous estimates, we get:

$||f||_{**} \leq ||f||_* + \sum_{j=1}^{k-1} C_j ||f||_* = (1 + \sum_{j=1}^{k-1} C_j) ||f||_*$

Let $C = 1 + \sum_{j=1}^{k-1} C_j$ . Then we have:

$||f||_{**} \leq C ||f||_*$

The Grand Finale: Equivalence Proven! :tada:

We've done it, guys! We've successfully shown that there exists a constant C such that $||f||_{**} \leq C||f||_*$ . Combining this with our earlier result that $||f||_* \leq ||f||_{**}$ , we've proven that the norms $||f||_*$ and $||f||_{**}$ are equivalent on $C^k[a, b]$ .

Why Does This Matter?

So, why did we go through all this trouble? Well, the equivalence of norms is a fundamental concept in functional analysis with significant implications. Here are a couple of key reasons why this result is important:

Convergence: Equivalent norms induce the same notion of convergence. This means that a sequence of functions converges in one norm if and only if it converges in the other norm. This can be extremely useful in simplifying proofs and calculations.
Completeness: Equivalent norms preserve completeness. If a space is complete with respect to one norm, it is also complete with respect to any equivalent norm. Completeness is a crucial property for many applications, such as solving differential equations.

Real-World Applications

The concepts we've explored today aren't just abstract mathematical ideas. They have practical applications in various fields, including:

Numerical Analysis: When approximating solutions to differential equations, understanding the equivalence of norms helps us choose appropriate numerical methods and analyze their convergence properties.
Signal Processing: In signal processing, functions represent signals, and norms are used to measure the energy or strength of a signal. Equivalent norms provide alternative ways to measure signal strength, which can be useful in different contexts.
Machine Learning: Norms play a crucial role in machine learning, particularly in regularization techniques. Understanding the equivalence of norms can help us design more effective regularization methods and improve the performance of machine learning models.

Conclusion: A Norm-al Day in Functional Analysis

Well, guys, that was quite a journey! We've delved into the world of $C^k[a, b]$ spaces, explored two different norms, and proven their equivalence using the powerful Taylor's Theorem. This is a testament to the beauty and interconnectedness of mathematical concepts.

Understanding the equivalence of norms is a crucial step in mastering functional analysis and its applications. I hope this guide has shed some light on this important topic and empowered you to explore further. Keep those mathematical gears turning, and I'll see you in the next adventure! :rocket: