Corollary of Banach Steinhaus theorem

Question

If $\{M_n\}_{n∈\mathbb{N}}$ is a family of continuous operators for $X$ Banach to $Y$ normed, such that $M_n(x)$ converges to $M(x)$ for all $x ∈ X$, then $M$ is a linear bounded operator and $||M||_{L(X,Y)}≤\liminf_{n→∞}||M_n||_{L(X,Y)}$. I cant understand of which set is taken the inf?

Answer 1

The $\liminf$ of a sequence $(a_n)$ of real numbers is defined as $$\liminf_{n\to \infty}a_n:=\lim_{n\to \infty}\inf_{k\geqslant n}a_k.$$ Here, this is justified because for each $x$ and each $n$, $$\lVert M_nx\rVert_Y\leqslant \lVert M_n\rVert_{L(X,Y)}\lVert x\rVert_X.$$ Taking the $\liminf_{n\to \infty}$ on both sides yields $$\lVert Mx\rVert_Y\leqslant \liminf_{n\to \infty}\lVert M_n\rVert_{L(X,Y)}\lVert x\rVert_X,$$ and we conclude the wanted result by the definition of the operator norm.

This gives a bound of the norm of the limit operator as a function of the norm of $M_n$'s.

It's worth pointing out that the sequence $(\lVert M_n\rVert_{L(X,Y)})_{n\geqslant 1}$ is not necessarily convergent: take $(e_n)_{n\geqslant 1}$ a Hilbert basis of $X=Y$ (assuming $X$ and $y$ are infinite-dimensional Hilbert spaces) and define $M_n(x):=\langle x,e_n\rangle e_n$: we have $M_n(x)\to 0$ for each $x$ and $\lVert M_n\rVert=1$ for each $n$.