Question

Find a Banach space \(X\) and a convex continuous bounded below function \(f\) on \(X\) such that \(\lim _{\|x\| \rightarrow \infty}(f(x))=\infty\) and \(f\) does not attain its infimum on \(X .\) Can \(X\) be reflexive? Can \(X\) be finite- dimensional if we drop the requirement on the limit of \(f ?\)

Short answer

Use \(X = C([0, 1])\) with \(f(x) = \int_{0}^{1} e^{x(t)}dt\). \(X\) is not reflexive, and it cannot be finite-dimensional without the limit condition.

Step-by-step solution

- Choose a Banach space

Select a specific Banach space. Consider using the space of continuous functions on a closed interval, e.g., let \(X = C([0, 1])\), the space of continuous functions on \([0, 1]\) with the sup norm.

- Define the function \(f\)

Define a convex, continuous, and bounded below function \(f: C([0, 1]) \rightarrow \textbf{R}\). For example, let \(f(x)= \int_{0}^{1} e^{x(t)}dt\) where \(x \in C([0, 1])\).

- Show the limit condition

Show that \lim_{\|x\| \rightarrow \infty} f(x) = \infty. Given the definition of \(f\), observe that as \|x\| \rightarrow \infty, some values of \(x(t)\) must grow very large, causing \(f(x)\) to grow exponentially.

- Evaluate the infimum

Consider the infimum of \(f\) on \(C([0, 1])\). Even though \(f\) is bounded below, \(f\) need not attain its infimum. This is because the minimum might be approached but never actually reached.

- Reflexivity

Determine whether \(X\) can be reflexive. For \(X = C([0, 1])\), it is known that this space is not reflexive, matching the conditions in the problem.

- Finite-dimensional case

Analyze whether \(X\) can be finite-dimensional if we drop the condition \lim_{\|x\| \rightarrow \infty} f(x) = \infty. For finite-dimensional spaces, every continuous function bounded below will attain its infimum, so \(X\) cannot be finite-dimensional without this condition.

Key concepts

Convex Continuous Function

A convex continuous function is a type of mathematical function that is both convex and continuous. To be convex, the function must satisfy the condition that for any two points on its graph, the line segment joining them lies above or on the graph. This can be more formally expressed as:

For all points, say, \(x \) and \(y \), in the domain of the function \(f\), and for any \( \theta \) in the interval [0, 1], the following must hold:
\[f(\theta x + (1-\theta)y) \leq \theta f(x) + (1-\theta) f(y)\].

In our example, the function \(f(x)= \int_{0}^{1} e^{x(t)}dt\) is convex because the exponential function and the integral both preserve convexity. Continuous means that there are no jumps, breaks, or holes in the graph of the function, which simplifies the comprehension of behavior within the function’s domain. This is vital for ensuring the function behaves consistently as \(x(t)\) ranges over \(C([0, 1]) \), which stands for the space of continuous functions over the interval \[0, 1\].

Reflexivity

The concept of reflexivity in functional analysis is crucial to understanding the properties of function spaces. A Banach space is reflexive if every bounded sequence has a weakly convergent subsequence. This means that the Banach space is essentially equivalent to its double dual space.

For the space \(X = C([0, 1])\), it is well-known that this space is not reflexive. One significant outcome of this non-reflexivity is that certain desirable properties, such as the attainment of the infimum of bounded functions, are not guaranteed. Since reflexive spaces have better structural properties, their absence here is key to why the infimum of \( f \) is not attained. Reflexivity ensures better control over sequences within the space, which we see is lacking in this case.

In simpler terms, reflexivity could have meant some kind of symmetry that helps in solving optimization problems easier, but the space \(C([0, 1]) \) does not have this property.

Finite-Dimensional Spaces

Finite-dimensional spaces are easier to handle compared to infinite-dimensional spaces. A space is finite-dimensional if it has a finite basis – a set of vectors that can be used to represent any element of the space as a linear combination of these vectors. The textbook example here tells us that if we drop the condition that \( \lim_{\|x\| \rightarrow \infty} f(x) = \infty \), then \(X \) can be finite-dimensional.

However, in finite-dimensional spaces, continuous functions that are bounded below, like \(f(x) \), would always attain their infimum because of compactness principles in finite dimensions. This means if there’s an infimum value within the range of a bounded below continuous function, there will be at least one point in the space where this infimum is achieved.

Thus, for space \(X \) to not attain its infimum point with function \(f \), \(ot\) finite-dimensional, implying that our space needs to be infinite-dimensional. Finite dimensions would not allow for such behavior because of structural limitations and compactness. This adds another layer of understanding to why certain mathematical properties hold in infinite dimensions but not in finite ones.