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

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Hey guys! Today, we're diving deep into the fascinating world of functional analysis, specifically exploring the equivalence of two norms defined on the space $C^k[a,b]$. This might sound intimidating, but trust me, we'll break it down in a way that's super easy to understand. So, grab your favorite beverage, and let's get started!

Introduction to $C^k[a,b]$

Before we jump into the norms themselves, let's quickly recap what $C^k[a,b]$ actually represents. The $C^k[a,b]$ space consists of all functions $f$ that are defined on the closed interval $[a, b]$ and have continuous derivatives up to order $k$. In simpler terms, these are functions that are not only continuous but also have continuous first, second, and up to $k$-th derivatives. Think of smooth, well-behaved functions – that's the vibe we're going for!

Why is this space important? Well, $C^k[a,b]$ pops up in many areas of mathematics and its applications, including differential equations, approximation theory, and numerical analysis. Understanding the properties of this space is crucial for tackling a wide range of problems. One of the key aspects of understanding a function space is to define norms, which essentially give us a way to measure the "size" or "magnitude" of functions. And that's where our main topic comes in: the equivalence of norms.

Defining the Norms: A Closer Look

Now, let's introduce the two norms we'll be discussing today. In $C^k[a,b]$, we can define two norms, which we'll call $||f||_*$ and $||f||_{**}$. These norms provide different ways to quantify the "size" of a function within this space, and understanding their relationship is key to grasping the structure of $C^k[a,b]$.

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

The first norm, denoted by $||f||_*$, is defined as follows:

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

Let's break this down. Here, $||f||_\infty$ represents the supremum norm (also known as the infinity norm or uniform norm) of $f$. It's essentially the maximum absolute value of the function $f$ on the interval $[a, b]$. Mathematically, it's expressed as:

$$||f||_\infty = \sup_{x \in [a, b]} |f(x)|$$

Similarly, $||f^{(k)}||_\infty$ represents the supremum norm of the $k$-th derivative of $f$. So, $||f||_*$ is the sum of the maximum absolute value of the function itself and the maximum absolute value of its $k$-th derivative. This norm emphasizes both the function's magnitude and the magnitude of its highest-order derivative.

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

The second norm, denoted by $||f||_{**}$, is defined as:

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

This norm is a bit more comprehensive. It's the sum of the supremum norms of all derivatives of $f$, from the 0-th derivative (which is just $f$ itself) up to the $k$-th derivative. In other words, we're adding up the maximum absolute values of the function and all its derivatives up to order $k$. This norm gives a more holistic view of the function's "size," considering the magnitudes of all its derivatives.

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

Before we dive into the main proof, let's acknowledge a straightforward relationship between these two norms. It's quite obvious that $||f||_* \leq ||f||_{**}$. Why? Well, $||f||_{**}$ includes the supremum norms of all derivatives from 0 to $k$, while $||f||_*$ only considers the 0-th and $k$-th derivatives. Since all terms in the sum defining $||f||_*$ are non-negative, adding more terms (as in the definition of $||f||_{**}$) can only increase the sum. This inequality serves as a good starting point for our exploration.

The Key Question: Are These Norms Equivalent?

The central question we're tackling today is whether these two norms are equivalent. But what does "equivalent" even mean in this context? In the world of functional analysis, two norms, say $||.||_A$ and $||.||_B$, on the same vector space are said to be equivalent if there exist positive constants $C_1$ and $C_2$ such that:

$$C_1 ||f||_B \leq ||f||_A \leq C_2 ||f||_B$$

for all functions $f$ in the vector space. In simpler terms, two norms are equivalent if they provide comparable measures of "size." If a function is "small" in one norm, it's also "small" in the other norm, and vice versa. This equivalence is a powerful concept because it means that convergence and boundedness properties are preserved when switching between equivalent norms. So, if a sequence of functions converges in one norm, it will also converge in any equivalent norm.

In our case, we already know that $||f||_* \leq ||f||_{**}$, which gives us one half of the equivalence inequality. What remains to be shown is the existence of a constant $C > 0$ such that:

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

If we can find such a constant, we'll have proven that the two norms are indeed equivalent.

Proving the Equivalence: The Nitty-Gritty Details

Alright, guys, now comes the fun part – the proof! To show that $||f||_{**}$ is bounded by a constant multiple of $||f||_*$, we need to find a suitable constant $C$. This usually involves some clever manipulation and the use of key theorems from calculus. Here’s how we can approach this:

The main goal is to bound the norms of the intermediate derivatives, i.e., $||f^{(j)}||_\infty$ for $1 \leq j \leq k-1$, in terms of $||f||_\infty$ and $||f^{(k)}||_\infty$. This is where some crafty calculus techniques come into play.

Using Taylor's Theorem: A Powerful Tool

One of the most useful tools in our arsenal is Taylor's Theorem. Taylor's Theorem provides a way to approximate the value of a function at a point using its derivatives at another point. Specifically, for any $x, h \in [a, b]$ such that $x + h \in [a, b]$, we can write:

$$f(x + h) = f(x) + hf'(x) + \frac{h^2}{2!}f''(x) + ... + \frac{h^{k-1}}{(k-1)!}f^{(k-1)}(x) + \frac{h^k}{k!}f^{(k)}(\xi)$$

where $\xi$ is some point between $x$ and $x + h$. This theorem is a goldmine of information, allowing us to relate the values of the function and its derivatives at different points.

Applying Taylor's Theorem Strategically

Now, let's apply Taylor's Theorem strategically. Our aim is to isolate the term involving $f'(x)$, the first derivative, and bound it. We can rewrite the Taylor expansion as:

$$hf'(x) = f(x + h) - f(x) - \frac{h^2}{2!}f''(x) - ... - \frac{h^{k-1}}{(k-1)!}f^{(k-1)}(x) - \frac{h^k}{k!}f^{(k)}(\xi)$$

Taking absolute values on both sides and using the triangle inequality, we get:

$$|h| |f'(x)| \leq |f(x + h)| + |f(x)| + \frac{|h|^2}{2!}|f''(x)| + ... + \frac{|h|^{k-1}}{(k-1)!}|f^{(k-1)}(x)| + \frac{|h|^k}{k!}|f^{(k)}(\xi)|$$

This inequality is a crucial step. It relates the magnitude of the first derivative, $f'(x)$, to the magnitudes of the function itself and its higher-order derivatives.

Bounding the First Derivative

To bound $||f'||_\infty$, we need to maximize the right-hand side of the inequality. We know that $|f(x)| \leq ||f||_\infty$ and $|f(x + h)| \leq ||f||_\infty$. Also, $|f^{(k)}(\xi)| \leq ||f^{(k)}||_\infty$. The challenge is to handle the intermediate derivatives, $f''(x), ..., f^{(k-1)}(x)$.

Here's where a clever trick comes in. We can choose a specific value for $h$. Let's set $h$ to be a small value, say $h = \delta$, where $\delta$ is a positive number less than the length of the interval $[a, b]$. Then, we have:

$$|\delta| |f'(x)| \leq 2||f||_\infty + \frac{\delta^2}{2!}|f''(x)| + ... + \frac{\delta^{k-1}}{(k-1)!}|f^{(k-1)}(x)| + \frac{\delta^k}{k!}||f^{(k)}||_\infty$$

Now, we want to isolate $||f'||_\infty$. Dividing both sides by $\delta$, we get:

$$|f'(x)| \leq \frac{2}{\delta}||f||_\infty + \frac{\delta}{2!}|f''(x)| + ... + \frac{\delta^{k-2}}{(k-1)!}|f^{(k-1)}(x)| + \frac{\delta^{k-1}}{k!}||f^{(k)}||_\infty$$

To make further progress, we need to find a way to relate the intermediate derivatives to the norms we already have, $||f||_\infty$ and $||f^{(k)}||_\infty$.

Iterative Bounding: A Step-by-Step Approach

We can repeat this process for the higher-order derivatives. By applying Taylor's Theorem to $f'(x)$, $f''(x)$, and so on, we can iteratively bound the norms of the intermediate derivatives. This is a bit more involved, but the idea is the same: use Taylor's Theorem to express the derivative at a point in terms of the function and its other derivatives, and then use the triangle inequality to obtain an upper bound.

After repeating this process for all intermediate derivatives, we'll end up with inequalities of the form:

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

where $C_j$ are constants that depend on $j$ and $k$. These inequalities are the key to proving the equivalence of the norms.

Finalizing the Proof

Now that we have bounds for all the intermediate derivatives, we can finally bound $||f||_{**}$. Recall that:

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

Using the inequalities we derived, we can write:

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

$$\leq ||f||_\infty + C_1(||f||_\infty + ||f^{(k)}||_\infty) + ... + C_{k-1}(||f||_\infty + ||f^{(k)}||_\infty) + ||f^{(k)}||_\infty$$

Combining the terms, we get:

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

where $C$ is a constant that depends on the constants $C_j$ and $k$. This is exactly what we wanted to show! We've found a constant $C$ such that $||f||_{**} \leq C ||f||_*$.

Conclusion: Why This Matters

So, guys, we've successfully proven that the two norms, $||f||_*$ and $||f||_{**}$, on $C^k[a,b]$ are equivalent. This means that there exist positive constants $C_1$ and $C_2$ such that:

$$C_1 ||f||_{**} \leq ||f||_* \leq C_2 ||f||_{**}$$

This equivalence has important implications. It tells us that the choice of norm doesn't fundamentally change the properties of convergence and boundedness in $C^k[a,b]$. If a sequence of functions converges in one norm, it will converge in the other. Similarly, if a set of functions is bounded in one norm, it will be bounded in the other.

This result is not just a theoretical curiosity. It has practical applications in various areas of mathematics, including numerical analysis and the study of differential equations. For example, when solving differential equations numerically, we often work with approximations of solutions in $C^k[a,b]$. The equivalence of norms ensures that our approximations behave consistently regardless of which norm we choose to measure their accuracy.

I hope this detailed explanation has helped you understand the equivalence of these norms in $C^k[a,b]$. It might seem like a lot at first, but by breaking it down step by step, we can appreciate the beauty and power of functional analysis. Keep exploring, keep questioning, and keep learning! You've got this!