Let X and Y be infinite-dimensional Banach spaces, and let T: X → Y be a linear map. Which statement is correct?
AT is automatically continuous, since linearity implies continuity in any normed space
BT is continuous if and only if it is bounded — that is, if there exists C such that ‖T(x)‖ ≤ C‖x‖ for all x
CT is continuous if and only if it maps Cauchy sequences to Cauchy sequences
DT cannot be continuous because infinite-dimensional spaces are not compact
In finite-dimensional spaces, every linear map is automatically continuous — this is a special theorem about finite dimensions. In infinite-dimensional spaces, this fails: you can construct linear maps that send bounded sequences to unbounded sequences. The correct statement is that, for linear maps between normed spaces, continuity and boundedness are equivalent conditions. The norm bound ‖T(x)‖ ≤ C‖x‖ is the precise condition that rules out pathological behavior. Option C is true for metric spaces generally but is not the standard formulation for linear maps.
Question 2 Multiple Choice
What is the geometric interpretation of the condition ‖T(x)‖ ≤ C‖x‖ for all x in the domain?
AT preserves angles between vectors — it maps orthogonal vectors to orthogonal vectors
BT maps the unit ball to a bounded set — the image of every bounded set is bounded
CT is an isometry — it preserves the norm of every vector
DT maps every vector to a vector of smaller norm
The condition ‖T(x)‖ ≤ C‖x‖ says that T cannot stretch vectors by more than a factor of C. Equivalently, the image of the unit ball {x : ‖x‖ ≤ 1} is contained in a ball of radius C in the output space — a bounded set. This is the geometric content of boundedness: T cannot send bounded inputs to unbounded outputs. An isometry (option C) would require ‖T(x)‖ = ‖x‖ exactly, which is much stronger. Option D is false — bounded operators can increase norms, just not without limit.
Question 3 True / False
For linear maps between normed spaces, continuity and boundedness are equivalent conditions.
TTrue
FFalse
Answer: True
This equivalence is one of the first fundamental theorems of functional analysis. If T is bounded (‖T(x)‖ ≤ C‖x‖), then for any convergent sequence xₙ → x: ‖T(xₙ) − T(x)‖ = ‖T(xₙ − x)‖ ≤ C‖xₙ − x‖ → 0, so T is continuous. Conversely, if T is continuous at 0, a simple argument shows T must be bounded. The key is that linearity links behavior at one point (0) to behavior everywhere, so local continuity implies global boundedness. This equivalence fails for nonlinear maps.
Question 4 True / False
Most linear map between infinite-dimensional Banach spaces is bounded.
TTrue
FFalse
Answer: False
This is the central misconception. In finite-dimensional spaces, the theorem that all linear maps are continuous holds because finite-dimensional spaces are topologically simple. In infinite-dimensional spaces, you can construct linear maps using a Hamel basis that are demonstrably unbounded — they send a sequence of unit vectors to vectors with norms growing without bound. Differential operators (like d/dx on function spaces) are the canonical examples from applications: they are linear but unbounded on natural function spaces. The distinction between bounded and unbounded operators is one of the main reasons functional analysis is non-trivial.
Question 5 Short Answer
Why does the proof that a bounded linear operator is continuous work, and why can this argument not be applied to show that an arbitrary linear map is continuous?
Think about your answer, then reveal below.
Model answer: If ‖T(x)‖ ≤ C‖x‖ for all x, then for any sequence xₙ → x: ‖T(xₙ) − T(x)‖ = ‖T(xₙ − x)‖ ≤ C‖xₙ − x‖ → 0. The bound C allows the norm of the input difference to control the norm of the output difference, giving continuity. For a general linear map without a uniform bound, the constant C might not exist or might be infinite — meaning for some sequence of inputs approaching 0, the outputs might not approach 0.
The whole point of the boundedness condition is that it provides the uniform constant C needed to make the epsilon-delta argument work globally. Without a uniform C, you might have ‖T(xₙ)‖ → ∞ even as ‖xₙ‖ → 0, which is exactly what happens for unbounded operators. Linearity alone maps 0 to 0 (T(0) = T(0+0) = 2T(0) implies T(0) = 0), but it does not control behavior near 0 uniformly across all directions in an infinite-dimensional space.