[proofplan]
Both conclusions follow from the [Uniform Boundedness Principle](/theorems/549). For part (1), SOT convergence gives pointwise boundedness of the sequence $(T_k)$, and the Uniform Boundedness Principle (applicable because $X$ is a Banach space) upgrades this to uniform boundedness in operator norm. For part (2), we use the lower semicontinuity of the norm under pointwise limits: for each $x$ in the unit ball, $\|Tx\|_Y = \lim_{k \to \infty} \|T_k x\|_Y \le \liminf_{k \to \infty} \|T_k\|_{\mathcal{L}(X,Y)} \|x\|_X$.
[/proofplan]
[step:Apply the Uniform Boundedness Principle to establish $\sup_k \|T_k\| < \infty$]
Since $T_k \to T$ in the SOT, for each $x \in X$ the sequence $(T_k x)_{k \in \mathbb{N}}$ converges in $Y$ and is therefore bounded:
\begin{align*}
\sup_{k \in \mathbb{N}} \|T_k x\|_Y < \infty \quad \text{for every } x \in X.
\end{align*}
This is precisely the pointwise boundedness hypothesis of the [Uniform Boundedness Principle](/theorems/549). The principle requires $X$ to be a Banach space (which is given) and $(T_k)$ to be a family of bounded linear operators from $X$ to the normed space $Y$ (which holds since each $T_k \in \mathcal{L}(X, Y)$). By the [Uniform Boundedness Principle](/theorems/549),
\begin{align*}
\sup_{k \in \mathbb{N}} \|T_k\|_{\mathcal{L}(X,Y)} < \infty.
\end{align*}
[guided]
The SOT convergence hypothesis states that for each fixed $x \in X$, the sequence $(T_k x)_{k \in \mathbb{N}}$ converges to $Tx$ in the norm of $Y$. Every convergent sequence in a normed space is bounded, so
\begin{align*}
\sup_{k \in \mathbb{N}} \|T_k x\|_Y < \infty \quad \text{for every } x \in X.
\end{align*}
This is a family of operators that is bounded *pointwise* — at each individual $x$, the norms $\|T_k x\|_Y$ stay finite, but the bound may depend on $x$. The question is whether the *operator norms* $\|T_k\|_{\mathcal{L}(X,Y)} = \sup_{\|x\| \le 1} \|T_k x\|_Y$ are also uniformly bounded.
This is exactly the gap that the [Uniform Boundedness Principle](/theorems/549) bridges. We verify its hypotheses:
- $X$ is a Banach space (given in the theorem statement). This is essential: the principle relies on the Baire Category Theorem, which requires completeness.
- $(T_k)_{k \in \mathbb{N}} \subset \mathcal{L}(X, Y)$ is a family of bounded linear operators into a normed space $Y$.
- The family is pointwise bounded, as established above.
By the [Uniform Boundedness Principle](/theorems/549), pointwise boundedness implies uniform boundedness:
\begin{align*}
\sup_{k \in \mathbb{N}} \|T_k\|_{\mathcal{L}(X,Y)} < \infty.
\end{align*}
The completeness of $X$ is consumed here. If $X$ were merely a normed space (not complete), the conclusion could fail: consider a dense subspace of $\ell^2$ on which one can construct a pointwise bounded family of functionals with unbounded norms.
[/guided]
[/step]
[step:Derive the $\liminf$ inequality for $\|T\|$]
Let $M := \sup_{k \in \mathbb{N}} \|T_k\|_{\mathcal{L}(X,Y)}$, which is finite by part (1). Fix $x \in X$ with $\|x\|_X \le 1$. Since $T_k x \to Tx$ in $Y$, the continuity of the norm gives
\begin{align*}
\|Tx\|_Y = \lim_{k \to \infty} \|T_k x\|_Y.
\end{align*}
For each $k$, the operator norm bound yields $\|T_k x\|_Y \le \|T_k\|_{\mathcal{L}(X,Y)} \|x\|_X \le \|T_k\|_{\mathcal{L}(X,Y)}$. Taking the $\liminf$ on both sides:
\begin{align*}
\|Tx\|_Y = \lim_{k \to \infty} \|T_k x\|_Y = \liminf_{k \to \infty} \|T_k x\|_Y \le \liminf_{k \to \infty} \|T_k\|_{\mathcal{L}(X,Y)}.
\end{align*}
The right-hand side is independent of $x$. Taking the supremum over all $x \in X$ with $\|x\|_X \le 1$:
\begin{align*}
\|T\|_{\mathcal{L}(X,Y)} = \sup_{\|x\|_X \le 1} \|Tx\|_Y \le \liminf_{k \to \infty} \|T_k\|_{\mathcal{L}(X,Y)}.
\end{align*}
[guided]
The inequality $\|T\| \le \liminf \|T_k\|$ states that the operator norm is *lower semicontinuous* with respect to SOT convergence. This is the infinite-dimensional analogue of the fact that limits can lose norm but never gain it.
Fix $x \in X$ with $\|x\|_X \le 1$. Since $T_k x \to Tx$ in $Y$, the norm function $\|\cdot\|_Y: Y \to \mathbb{R}$ (which is continuous) gives
\begin{align*}
\|Tx\|_Y = \lim_{k \to \infty} \|T_k x\|_Y.
\end{align*}
Now we use the operator norm bound: for each individual $k$,
\begin{align*}
\|T_k x\|_Y \le \|T_k\|_{\mathcal{L}(X,Y)} \|x\|_X \le \|T_k\|_{\mathcal{L}(X,Y)}.
\end{align*}
The $\liminf$ preserves this inequality (if $a_k \le b_k$ for all $k$, then $\liminf a_k \le \liminf b_k$):
\begin{align*}
\|Tx\|_Y = \lim_{k \to \infty} \|T_k x\|_Y = \liminf_{k \to \infty} \|T_k x\|_Y \le \liminf_{k \to \infty} \|T_k\|_{\mathcal{L}(X,Y)}.
\end{align*}
The right-hand side does not depend on $x$. Since this holds for every $x$ in the closed unit ball, taking the supremum over all such $x$ yields
\begin{align*}
\|T\|_{\mathcal{L}(X,Y)} = \sup_{\|x\|_X \le 1} \|Tx\|_Y \le \liminf_{k \to \infty} \|T_k\|_{\mathcal{L}(X,Y)}.
\end{align*}
The inequality can be strict. For instance, let $H = \ell^2$ and $T_k x = (0, \ldots, 0, x_k, 0, \ldots)$ (placing $x_k$ in the $k$-th coordinate). Then $\|T_k\| = 1$ for all $k$, but $T_k \to 0$ in the SOT, so $\|T\| = 0 < 1 = \liminf \|T_k\|$.
[/guided]
[/step]
