Why doesn't pointwise bounded imply uniform bounded?

real-analysis analysis

Solution 1:

The short answer is: there need not be a real number that is the supremum of the values of $\phi(x)$. You may have $\sup\{\phi(x)\mid x \in E\} = \infty$. If that is the case, you're out of luck.

Solution 2:

This an excellent example of a subtle issue that keeps coming up. The property of being pointwise bounded ensures that for every element of your domain of definition the sequence of functions is bounded at that point. Where as uniform boundedness says that there exists an upper bound that holds for every element of your domain.

I decided to answer this old question because I had a very similar thought regarding The Principle of Uniform Boundedness in my Functional Analysis class.
Another point worth mentioning is that this is yet another case where the quantifiers do not commute. Another classical example being Uniform Convergence vs Pointwise Convergence

Related

Permutation with Duplicates
Is the group isomorphism problem decidable for abelian groups?
Unbounded subset of $\mathbb{R}$ with positive Lebesgue outer measure
Is any divergence-free curl-free vector field necessarily constant?
Defining Measures on a Manifold: How To
How much rigour in proofs?
Semidirect Products in Category Theory?
Orbit-stabilizer theorem for Lie groups?
Do you need real analysis to understand complex analysis?
show that a subset of $\mathbb{R}$ is compact iff it is closed and bounded
How does this equality on vertices in the complex plane imply they are vertices of an equilateral triangle?
Proof of $H^k(X,\mathbf k) = H^k(X,\mathbb Z) \otimes \mathbf k$