Elementary Proof Of The Uniform Boundedness Theorem

Arxiv: Alan D. Sokal Uniform boundedness theorem. $X$ a Banach space $Y$ a normed linear space $ℱ$ a family of bounded linear operators from $X$ to $Y$. If $ℱ$ is pointwise bounded (i.e., $\sup_{T ∈ ℱ} \|Tx\| < ∞$ for all $x ∈ X$), then $ℱ$ is norm-bounded (i.e., $\sup_{T ∈ ℱ} \|T\| < ∞$). Lemma. Let $T$ be a bounded linear operator from a normed linear space $X$ to a normed linear space $Y$. Then for any $x ∈ X$ and $r > 0$, we have \[ \sup\limits_{x' ∈ B(x,r)} \| Tx' \| ≥ \|T\| r , \] where $B(x,r) = \{x' ∈ X \colon\: \|x'-x\| < r \}$. Proof. For $ξ ∈ X$ we have \[ \max\{ \| T(x+ξ) \| , \| T(x-ξ) \| \} ≥ \frac12 [ \| T(x+ξ) \| + \| T(x-ξ) \| ] ≥ \| T ξ \| , \] where the second $≥$ uses the triangle inequality in the form $\| α-β \| ≤ \|α\| + \|β\|$. Now take the supremum over $ξ ∈ B(0,r)$. ∎ Proof of the uniform boundedness theorem. Suppose $ℱ$ is not norm-bounded, i.e. $\sup_{T ∈ ℱ} \|T\| = ∞$, and choose $(T_n)_{n=1}^∞$ in $ℱ$ such that $\|T_n\| ≥ 4^n$. Then set $x_0 = 0$, and for $n ≥ 1$ use the lemma to choose inductively $x_n ∈ X$ such that $\| x_n - x_{n-1} \| ≤ 3^{-n}$ and $\| T_n x_n \| ≥ \frac{2}{3} 3^{-n} \| T_n \|$. The sequence $(x_n)$ is Cauchy, hence convergent to some $x ∈ X$; and it is easy to see that $\| x-x_n \| ≤\sum_{i=n+1}^{∞} 3^{-i}=\frac{1}{2} 3^{-n}$ and hence $\|T_nx-T_nx_n\|≤\frac{1}{2} 3^{-n}$. By triangle inequality \[\| T_n x \| ≥\|T_nx_n\|-\|T_nx-T_nx_n\|≥\frac{1}{6} 3^{-n} \| T_n \| ≥ \frac{1}{6} (4/3)^n → ∞ \] so ℱ is not pointwise bounded. ∎