Question

On the vector space \(\mathcal{F}^{\prime}[a, b]\) of complex continuous differentiable functions on the, interval \([a, b]\), set $$ \langle f| g)=\int_{a}^{b} \overline{f^{\prime}(x)} g^{\prime}(x) \mathrm{dr} \text { where } f^{\prime}=\frac{\mathrm{d} f}{\mathrm{~d} x}, \quad g^{\prime}=\frac{\mathrm{d} g}{\mathrm{~d} x} $$ Show that this is not an inner product, but becomes one if restricted to the space of functions \(f \in\) \(F^{\prime}[a, b]\) having \(f(c)=0\) for seme fixed \(a \leq c \leq b\). Is it a Hilbert space? Give a similar analysis for the case \(a=-\infty, b=\infty\), and restricting functions to those of compact support.

Short answer

The given equation does not establish an inner product on the vector space of complex continuous differentiable functions due to non-compliance with the positive-definiteness requirement. However, it becomes one if it's limited to the functions that pass through a certain point in the interval. Whether these spaces with the defined inner product form a Hilbert space cannot be determined without additional information. Similarly, for the case extending from negative to positive infinity, we can follow the same steps, but determining whether it forms a Hilbert space remains inconclusive without additional data.

Step-by-step solution

Proving it's not an inner product

The given equation is stated to be an inner product. To prove otherwise, we need to show that it does not satisfy at least one of the inner product axioms: (1) Conjugate Symmetry, (2) Linearity in the first argument, and (3) Positive-definiteness (a function with itself should not equal zero unless the function is zero). Consider the case of a simplest function \(f(x)=1\) on \(F'[a, b]\). When \(f\) and its conjugate is inserted into the equation, the result is \(0\), which violates the positive-definiteness axiom. Therefore, the given equation is not an inner product.

Proving it becomes an inner product

The problem specifies that the functions \(f \in F'[a, b]\) have \(f(c)=0\) for some fixed \(a ≤ c ≤ b\). This condition makes each function have at least one root in the interval, let's say at \(f(c)\). Now we try again with the function from the previous step, now passing through \(f(c)\). As before, the result is \(0), but now the function is zero at at least one point. It does not contradict the positive-definitiveness axiom, hence proving, under this limitation, the given equation becomes an inner product.

Checking for Hilbert space

A Hilbert space is a complete inner product space, meaning every Cauchy sequence in this space converges to a point within the space. However, as this problem does not provide enough information to determine convergence of sequences in this set, we cannot make a conclusion about whether it is a Hilbert space.

Analyzing the case from negative to positive infinity

When the interval extends to negative to positive infinity, and the subspace is restricted to functions of compact support (i.e., functions that are nonzero only on a compact subset of the interval), both steps 1 and 2 can be repeated. The restriction ensures that the functions still meet the limitations required for the equation to be an inner product. However, similar to step 3, we cannot make a conclusion about whether it shapes a Hilbert space without more information on the convergence of sequences within the space.

Key concepts

Inner Product Space

An inner product space is a vector space equipped with an additional structure, specifically an inner product. The inner product is a function that assigns a complex number to each pair of vectors and satisfies several axioms. These axioms are:

• Conjugate Symmetry: The inner product is symmetric under conjugation, meaning \(\langle f | g \rangle = \overline{\langle g | f \rangle}\).
• Linearity in the First Argument: For functions \(f\), \(g\), and \(h\), and scalars \(a\) and \(b\), \(\langle a f + b g | h \rangle = a \langle f | h \rangle + b \langle g | h \rangle\).
• Positive-Definiteness: For any vector \(f\), \(\langle f | f \rangle \geq 0\) with equality if and only if \(f\) is the zero vector.

The given equation may fail positive-definiteness if a function doesn't meet certain conditions, like not always being zero at some point on the interval. When dealing with functions of continuous support or bounded by certain conditions (such as zeros at fixed points), equations meeting these axioms constitute inner product spaces. Understanding this helps in bridging different mathematical concepts within vector analysis.

Continuous Differentiable Functions

Continuous differentiable functions are functions that have continuous derivatives. In simpler terms, their rate of change is smooth and there aren’t any abrupt changes. Differentiability is closely tied with continuity; if a function is differentiable at a point, it is continuous there. When considering the space \((\mathcal{F}'[a, b])\), we deal with functions that are not only defined on an interval but also have derivatives that remain continuous over that interval.
This makes these functions particularly useful in analysis and geometry, as they naturally allow transformation and provide consistent results when calculating gradients or doing integrals.In exercises with differentiable functions, ensuring continuity at specific points can change properties of mathematical constructs they are part of, such as transforming a non-inner product space into an inner product space by enforcing zero at a point.

Compact Support

Compact support in functions relates to functions that are essentially zero outside a certain subset of their domain. This means the function is only "active" or nonzero in a confined, bounded interval called the "support." Conceptually, you can think of it as having all its essential values condensed into a compact region.When functions are restricted to compact support, they behave well with respect to limits and integrations over infinite domains, often simplifying complex problems. In some analyses, especially those involving functions over the whole real line \([-\infty, \infty]\), focusing on functions of compact support ensures manageable and finite behavior when calculating integrals or solving differential equations. For example, when a function's activity is restricted to ensure positivity within its support region, this constraint helps enforce the positive-definiteness condition required by inner product spaces.

Positive-Definiteness Axiom

The positive-definiteness axiom is crucial for identifying inner products. It ensures that the inner product of a function with itself is non-negative and is zero only if the function itself is identically zero.In mathematical terms, for a function \(f\), the axiom requires \(\langle f | f \rangle \geq 0\) and \(\langle f | f \rangle = 0\) only if \(f = 0\). This concept guarantees that the inner product provides a kind of "length" or "magnitude" for functions in the space.In the context of the given exercise, failing to satisfy positive-definiteness means the supposed inner product setup doesn’t meet all necessary properties to be considered valid. Adjustments, like setting specific zeros or confining support, are techniques to overcome these hurdles, making certain mathematical structures valid under tighter conditions.