If Bounded Operators Converge Pointwise, Are Their Norms Bounded?

Suppose $(T_n)$ is a sequence of bounded linear operators from a Banach space $X$ into a normed space $Y$. If $T_nx$ converges in $Y$ for every $x\in X$, then the operator norms are uniformly bounded:

$$\sup_n \|T_n\| < \infty.$$

This is an application of the uniform boundedness principle, also called the Banach–Steinhaus theorem. The key assumption is that the domain $X$ is complete. The codomain need not be complete.

Why pointwise convergence is enough

For each fixed $x$, convergence of the sequence $(T_nx)$ implies that its values are bounded. Thus

$$\sup_n \|T_nx\| < \infty \qquad\text{for every }x\in X.$$

This is pointwise boundedness. The uniform boundedness principle says that for bounded linear operators on a Banach space, pointwise boundedness forces a single bound that works for every operator norm. The bound does not depend on $x$ or on $n$.

The statement is about boundedness of the sequence in operator norm; it does not say that $T_n$ converges in operator norm.

Proof using the Baire category theorem

For each positive integer $m$, define

$$E_m=\{x\in X: \|T_nx\|\le m\text{ for every }n\}.$$

Each $E_m$ is closed, since it is the intersection over $n$ of the closed sets $\{x:\|T_nx\|\le m\}$. Pointwise boundedness also gives $X=\bigcup_{m=1}^{\infty}E_m$: for any fixed $x$, some integer $m$ bounds all the values $\|T_nx\|$.

Because $X$ is Banach, the Baire category theorem implies that at least one $E_m$ has nonempty interior. So for some $x_0\in X$ and $r>0$, the open ball $B(x_0,r)$ lies in $E_m$. Since $x_0$ itself is in $E_m$, whenever $\|h\|<r$ we have

$$\|T_nh\|\le \|T_n(x_0+h)\|+\|T_nx_0\|\le 2m$$

for every $n$. Take any $u$ with $\|u\|=1$ and put $h=(r/2)u$. Then $\|h\|<r$, so

$$\|T_nu\|\le \frac{4m}{r}$$

for every $n$ and every unit vector $u$. Taking the supremum over unit vectors gives $\|T_n\|\le 4m/r$ for all $n$. Hence the operator norms are uniformly bounded.

A pointwise limit is also bounded

If $T_nx$ converges to $Tx$ for every $x$, then the limit map $T$ is linear, because each $T_n$ is linear and limits preserve addition and scalar multiplication. The uniform bound also shows that $T$ is bounded. If $\|T_n\|\le C$ for every $n$, then

$$\|Tx\|=\lim_{n\to\infty}\|T_nx\|\le C\|x\|.$$

Thus $T$ is a bounded linear operator and $\|T\|\le C$.

Example: pointwise convergence does not imply norm convergence

On the Hilbert space $\ell^2$, let $P_n$ keep the first $n$ coordinates and set all later coordinates to zero. For any $x=(x_1,x_2,\ldots)\in\ell^2$, the sequence $P_nx$ converges to $x$ in $\ell^2$, since the tail of a square-summable sequence tends to zero. Also, $\|P_n\|=1$ for every $n$.

Nevertheless, $P_n$ does not converge to the identity operator in operator norm. For every $n$, the operator $I-P_n$ keeps the tail, and applying it to the unit vector with a $1$ in coordinate $n+1$ gives a vector of norm $1$. Therefore $\|I-P_n\|=1$ for every $n$. This illustrates the distinction: the norms of the operators stay bounded, but pointwise convergence need not be uniform over the unit ball.

Why completeness matters

Without a Banach domain, the conclusion can fail. Let $X=c_{00}$, the space of finitely supported sequences, with the $\ell^2$ norm. Define $T_n:X\to\mathbb{R}$ by $T_n(x)=n x_n$. Each $T_n$ is bounded and has norm $n$, since the coordinate functional $x\mapsto x_n$ has norm $1$. Yet for every $x\in c_{00}$, $T_n(x)=0$ for all sufficiently large $n$, so $T_n(x)\to0$. The operator norms are unbounded because $c_{00}$ is not complete in the $\ell^2$ norm.

Common mistakes

• Omitting the Banach-space assumption: Pointwise convergence alone is not enough on an arbitrary normed domain.
• Confusing pointwise and norm convergence: Pointwise convergence concerns each fixed input; operator-norm convergence requires uniform control over the unit ball.
• Assuming the limit is bounded without justification: Once the uniform bound is established, passing to the pointwise limit proves boundedness of the limit operator.