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Understanding differentiable functions is crucial when verifying if a subset is a subspace within a vector space. A differentiable function is one that has a derivative at every point in its domain. This means that the function smoothly changes without any sudden jumps or breaks.
In the context of the given exercise, functions that are differentiable on the interval \[0,1\] form the set \(W\). This means that every function within \(W\) must be capable of being differentiated throughout the entire interval \[0,1\]. Differentiability implies continuity, as a function must be continuous to have a derivative. Therefore, any differentiable function on this interval is also continuous, making it a member of the broader set \(V\), which contains all continuous functions on the same interval.

A vector space is a collection of vectors, which can be thought of as points, that can be added together or scaled by numbers called scalars. Within vector spaces, there are specific operations and properties that must be satisfied. Simple examples include the vector space of all real numbers and the vector space of continuous functions.
The given exercise focuses on the vector space \(V\), which is the set of all continuous functions on the interval \[0,1\]. This means that any operation applicable to continuous functions, like addition or scalar multiplication, is valid across the entire set \(V\). For \(W\) to qualify as a subspace of \(V\), it must adhere to the properties of a vector space such as containing the zero vector, and closure under addition and scalar multiplication, all within its specified set of differentiable functions.

Closure properties are essential features of any subspace within a vector space. They ensure that certain operations, when applied to elements of the subspace, result in an element that is still within the subspace. For a subset to be a subspace, it must have closure under two operations: addition and scalar multiplication.

• Closure under addition: If you take any two functions \(f\) and \(g\) that are members of \(W\), their sum \(f+g\) must also be a function in \(W\). This is true here because the sum of two differentiable functions is still differentiable.
• Closure under scalar multiplication: This property holds if, for any function \(f\) in \(W\) and any scalar \(c\), the product \(c \cdot f\) is also in \(W\). In this scenario, multiplying a differentiable function by a scalar results in another differentiable function, asserting this closure property in \(W\).

These closures are foundational in proving that \(W\) is a subspace of \(V\).

The zero vector is a fundamental element in any vector space, including subspaces. It serves as the 'no effect' element in a space, where adding it to any vector doesn't alter that vector.
In function spaces, this zero vector is the zero function, which assigns a value of zero to every point in its domain.
Within this exercise, the zero vector is the function \(0(x) = 0\) for all \([0,1]\). This zero function is inherently differentiable as it remains constant, and its derivative is also zero.
Having the zero vector in \(W\) means the set fulfills one of the critical conditions for being a subspace. The presence of the zero function confirms that \(W\) can't exclude any essential component of the vector space \(V\), supporting \(W\)'s subspace characterization.