$T_n \rightarrow T$ then we have $||T|| \le \liminf(||T_n||)$ I know how to show that a cauchy sequence of linear continuous operators $T_n:X \rightarrow Y$ has a limit that is also such an operator(if Y is a Banach space), but I found this relation here too $||T|| \le \liminf(||T_n||)$ and I just don't know where this liminf comes into play. Does anybody here know?
 A: If $(T_n)$ is a Cauchy sequence converging to $T$ (with respect to the operator norm), then we have
$$\lVert T\rVert = \lim_{n\to\infty} \lVert T_n\rVert$$
by the continuity of the norm ($\left\lvert \lVert T\rVert - \lVert S\rVert\right\rvert \leqslant \lVert T-S\rVert$).
You get the $\liminf$ if the convergence is in a different sense, e.g. pointwise convergence (although of course it remains correct in the case of norm convergence, we just have something more precise then).
If $(T_n)$ is a sequence of continuous linear operators $X\to Y$, where $X$ and $Y$ are normed spaces that converges pointwise to a linear operator $T$, and if $M = \liminf\limits_{n\to\infty} \lVert T_n\rVert < \infty$ (if $X$ is a Banach space, a pointwise convergent sequence is uniformly bounded by the Banach-Steinhaus theorem, so the condition is automatically fulfilled; for incomplete $X$, it must be explicitly stated), then $T$ is continuous, and
$$\lVert T\rVert \leqslant M.$$
For, let $x \in X$ with $\lVert x\rVert \leqslant 1$. For every $\varepsilon > 0$, there are infinitely many $n_k$ with $\lVert T_{n_k}\rVert < M + \varepsilon$. Since the subsequence $(T_{n_k})$ also converges pointwise to $T$, we have
$$Tx = \lim_{k\to\infty} T_{n_k}x.$$
But $\lVert T_{n_k}x\rVert < M+\varepsilon$ for all $k$, and hence $\lVert Tx\rVert \leqslant M+\varepsilon$. That holds for all $x$ with $\lVert x\rVert \leqslant 1$, so
$$\lVert T\rVert = \sup_{\lVert x\rVert \leqslant 1} \lVert Tx\rVert \leqslant M + \varepsilon.$$
That holds for all $\varepsilon > 0$, hence $\lVert T\rVert \leqslant M$.
