# Is the uniform boundedness principle not trivially obvious?

I read that the uniform boundedness principle is one of the big theorems in functional analysis. However after looking at it, I'm not sure what is so significant about it, it seems trivially obvious.

The uniform boundedness principle states that if we have a family of operators $F = (T_n)$ from a Banch space $X$ to a normed space $Y$ that are pointwise bounded, then they are uniformly bounded in the operator norm.

This seems trivial, if the family of operators is pointwise bounded then we know that for any $x$ we take in the domain $X$, the resulting value after applying one of the operators from $F$ to it will also be a finite value. So clearly the operator norm (which accounts for all $x \in X$ and $T_n \in F$) will also be finite.

Am I missing something here, why is this theorem so significant and considered an important result?

• Hint: imagine a hypothetical scenario where $x_k$ is a sequence of points in $X$ such that $|| x_k || = 1$ for all $k \in \mathbb N$, and $\sup_{T \in F}||T(x_k)|| = k$ for each $k$. – Kenny Wong Mar 18 '17 at 11:12
• @KennyWong Ok so for any $x_k$ we have $\sup_{T in F} ||T(x_k)|| = k$ and therefore as $k$ can be arbitrarily large $\sup_{T in F} ||T(x_k)|| = \infty$. So in your example we don't have pointwise boundedness, and thus we can't apply the principle of uniform boundedness. I don't think I understand what your example is supposed to illustrate? – ManUtdBloke Mar 18 '17 at 11:37
• For any particular fixed $x_k$, we have $\sup_{T \in F} || T(x_k) || = k$. As I said, $k$ is fixed. Now, $k$ is a finite number. So $\sup_{T \in F} || T(x) ||$ is finite at the point $x= x_k$, for any choice of $x_k$. – Kenny Wong Mar 18 '17 at 11:43
• However, $\sup_{T \in F} || T || = \sup_{T \in F} \left( \sup_{|| x|| = 1} || T(x) || \right) = \infty$. – Kenny Wong Mar 18 '17 at 11:44
• Excellent. The principle of uniform boundedness proves that my hypothetical example can never occur! – Kenny Wong Mar 18 '17 at 12:16

If you have a family of operators from a normed space to a normed space that are pointwise bounded then they need not be uniformly bounded. It is crucial that the domain is complete (and one uses some version of Baire's category theorem in the proof). But I suppose a concrete example is better. Let $$V=\{ x=(x_k)_{k\geq 1} : \sup_{k\geq 1} k |x_k| < +\infty \}$$ and equip $V$ with the uniform norm: $\|x\| = \sup_{k\geq 1} |x_k|$. Then $V$ is not a Banach space (not complete). On the other hand, the family $F_k(x) = k x_k$, $k\geq 1$ is a family of pointwise bounded operators from $V$ to ${\Bbb R}$: By definition of $V$, for every $x\in V$ there is $c(x)<+\infty$ so that $$|F_k(x)| \leq c(x), \ \ k\geq 1$$ so the family is indeed point wise bounded. However, $$\|F_k\| = k$$ (take an $x$ with $1$ in the $k$'th place, zero elsewhere) so it is not uniformly bounded.