Question

$$ \text { Show that } C[0,1] \oplus C[0,1] \text { is isomorphic to } C[0,1] \text { . } $$

Short answer

The map \( \Phi(f,g)(x) = f(x) + g(x) \) establishes an isomorphism between \(C[0,1] \oplus C[0,1]\) and \(C[0,1]\).

Step-by-step solution

- Understand the Notation

The notation \(C[0,1]\) represents the set of continuous functions on the interval \([0,1]\). \( \oplus \) denotes the direct sum (or direct product) of these spaces.

- Define the Direct Sum

\(C[0,1] \oplus C[0,1]\) refers to the set of ordered pairs \((f,g)\) where \(f\) and \(g\) are both continuous functions on \([0,1]\). This means each element in \(C[0,1] \oplus C[0,1]\) is a pair of continuous functions.

- Define an Isomorphism

An isomorphism between two spaces means there exists a bijective linear map that preserves the structure. We must find such a map between \(C[0,1] \oplus C[0,1]\) and \(C[0,1]\).

- Construct the Isomorphism Map

Consider the map \(\Phi: C[0,1] \oplus C[0,1] \to C[0,1] \) defined by \( \Phi(f,g)(x) = f(x) + g(x) \). This map takes a pair of continuous functions \((f,g)\) and maps them to their sum.

- Verify Linearity

Check that \( \Phi \) is linear: For \( (f,g), (h,k) \in C[0,1] \oplus C[0,1] \) and scalars \(a\) and \(b\), \( \Phi(a(f,g) + b(h,k)) = \Phi((af+bh, ag+bk)) = af + ag + bh + bk = a(f+g) + b(h+k) = a \Phi(f,g) + b \Phi(h,k) \).

- Verify Injectivity

\( \Phi \) is injective: If \( \Phi(f,g) = \Phi(h,k)\), then \(f(x) + g(x) = h(x) + k(x)\) for all \(x \in [0,1]\). If \( \Phi(f,g) = 0\), then \(f(x) + g(x) = 0\) for all \(x \in [0,1]\) implying \(f = -g\).

- Verify Surjectivity

\( \Phi \) is surjective: For any \(u \in C[0,1]\), let \(f(x) = \frac{u(x)}{2}\) and \(g(x) = \frac{u(x)}{2}\). Since \(u(x) \in C[0,1]\), \(f\) and \(g\) are also in \(C[0,1]\). Then, \( \Phi(f,g) = f(x) + g(x) = \frac{u(x)}{2} + \frac{u(x)}{2} = u(x) \).

- Conclusion

Since \( \Phi \) is a bijective linear map, it shows that \(C[0,1] \oplus C[0,1]\) is isomorphic to \(C[0,1]\).

Key concepts

Continuous Functions

Let's start with the basics. A function is continuous if, intuitively, you can draw its graph without lifting your pencil off the paper. More formally, a function \(f: [0, 1] \rightarrow \mathbb{R}\) is continuous if for every point \(x_0\) in the interval \[0, 1\], and for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(|x - x_0| < \delta\), we have \(|f(x) - f(x_0)| < \epsilon\).
This might sound technical, but it's just a way to ensure that small changes in the input (x) result in small changes in the output (f(x)), making the function predictable and smooth.
Now, when we talk about \(C[0,1]\), we refer to the set of all continuous functions defined on the interval \[0, 1\]. This set is quite broad and can include a variety of functions, such as polynomials, trigonometric functions, and exponential functions, as long as they are continuous on \[0, 1\].

Direct Sum

The direct sum, denoted by \( \oplus \), is a concept used to combine two mathematical objects, such as vector spaces or modules, into a larger one. For example, if we take two continuous function spaces \(C[0,1]\), their direct sum \(C[0,1] \oplus C[0,1] \) is the set of ordered pairs of functions.
This means that every element in \(C[0,1] \oplus C[0,1] \) is a pair of continuous functions, say \( (f, g) \), where both \(f \) and \ g \ are continuous functions on \[0,1\]. The direct sum allows us to work with pairs of functions as single entities.
Here’s why it’s useful: it allows us to generalize problems and make more complex spaces easier to work with by combining simpler ones. In practice, if each vector space (or function space) has certain properties, their direct sum will inherit these properties, making the combination as manageable as the individual spaces.

Linear Map

A linear map (or linear transformation) is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. In other words, a linear map \(T\) from a vector space \V\ to another vector space \W\ satisfies: \[ T(av + bw) = aT(v) + bT(w) \] for any vectors \(v, w \) in \V\ and any scalars \(a, b \).
Linear maps are crucial because they maintain the structure of the spaces they map. In the context of our exercise, the linear map \( \Phi: C[0,1] \oplus C[0,1] \rightarrow C[0,1] \) defined by \( \Phi(f,g)(x) = f(x) + g(x) \) is a perfect example. This map adds the two functions from the direct sum to produce a single continuous function.
Verifying properties like linearity, injectivity, and surjectivity ensures that \(\Phi\) not only preserves the structure but also forms a perfect one-to-one correspondence (isomorphism) between \(C[0,1] \oplus C[0,1]\) and \(C[0,1]\). This is why understanding linear maps is key to understanding more complex structures in functional analysis.