Let $(V,\|\cdot\|_V)$ and $(W,\|\cdot\|_W)$ be normed vector spaces. By $L(V,W)$ we denote the space of bounded linear operators from $V$ to $W.$ If $W$ is complete, then $L(V,W)$ with the operator norm $\|T\| := \sup \left\{\|Tu\|_W : \|u\|_V \leq 1\right\}$ is a Banach space.
Now assume in addition that $\gamma$ is probability measure on $V$ with $\mathrm{supp}\, \gamma = V$ and with existing first moment.
My Question: Is $L(V,\mathbb{R})$ closed subspace in $(\mathrm{L}_1(V),\|\cdot\|_1)$?
In this case $L(V,\mathbb{R})$ with $\|\cdot\|_1$ would also be a Banach space.
The answer is no. Consider $V=l^2(\mathbb{N}, \mathbb{R})$ and let $e_i$ be the i-th basis vector of the usual Schauder basis. Set $$ \gamma = \sum_{n\geq 1} 2^{-n} \delta_{e_n}$$ And consider the following sequence of bounded linear functionals $$l_m (x) = \sum_{j=1}^m j \cdot x_j $$ This forms a Cauchy sequence as for $m>k$ holds $$ \Vert l_m - l_k \Vert = \sum_{j=k+1}^m j\cdot 2^{-j}$$ I leave it to you to you to show that this sequence does not converge to some bounded operator (assume it does and then do the computation for the norm to get a contradiction).
Added: I forgot to check that my probability measure admits the first moment (I did not even know how it was defined, thanks to H17 for telling me). In fact not only does the first moment exist, but all moments do. Indeed, for $k\geq 1$ we have $$ \int \Vert x \Vert^k \gamma(dx) = \sum_{n\geq 1} \Vert e_n \Vert^k \cdot 2^{-n} = \sum_{n\geq 1} 2^{-n} = 1. $$