Here is an example where Jensen’s inequality $f(E[\mathbf{X}])\leq E[f(\mathbf{X})]$ for convex functions $f$ fails when $f$ is defined over an infinite dimensional vector space, even when all expectations are finite.
Define $\mathcal{X}$ as the set of all infinite sequences $\mathbf{x}=\{x_i\}_{i=1}^{\infty}$ such that
$\lim_{i\rightarrow\infty} x_i2^i$ exists and is a real number. Define
$f:\mathcal{X}\rightarrow\mathbb{R}$ by
$$ f(\mathbf{x}) = \lim_{i\rightarrow\infty} x_i2^i $$
It is easy to verify that $\mathcal{X}$ is a convex set and $f$ is a convex function.
For each $m \in \{1, 2, 3, ...\}$ define $\mathbf{e}^{(m)}$ as the infinite sequence that is 1 in entry $m$ and zero in all other entries. Then $f(\mathbf{e}^{(m)}) = 0$ for all $m \in \{1, 2, 3, ...\}$. Let $G$ be a random variable with mass function $P[G=m]=(1/2)^m$ for $m \in \{1, 2, 3, …\}$. Define the random sequence $\mathbf{X}= \mathbf{e}^{(G)}$, so
$$ \mathbf{X} = \left\{ \begin{array}{ll}
\mathbf{e}^{(1)}=\{1, 0, 0, 0, ...\} &\mbox{ with prob 1/2} \\
\mathbf{e}^{(2)}=\{0, 1, 0, 0, ...\} &\mbox{ with prob 1/4} \\
\mathbf{e}^{(3)}=\{0, 0, 1, 0, ...\} &\mbox{ with prob 1/8} \\
\cdots
\end{array}
\right.$$
Then $f(\mathbf{X}) = 0$ always, and so $E[f(\mathbf{X})]=0$.
However:
$$ E[\mathbf{X}] = \{1/2, 1/4, 1/8, ...\} =\{2^{-i}\}_{i=1}^{\infty} \implies f(E[\mathbf{X}])=1$$
$\Box$
The reason the above (counter)example exists is because the set $\mathcal{X}$ is not closed and/or the function $f$ is not continuous. Let $V$ be a normed vector space (possibly infinite dimensional). Here is a statement and proof of Jensen's inequality for this case.
Claim (Infinite dimensional Jensen):
Suppose $\mathcal{X}$ is a closed and convex subset of $V$ and
$f:\mathcal{X}\rightarrow\mathbb{R}$ is a continuous and convex function. Let
$X$ be a random vector that takes values in $\mathcal{X}$ and assume $E[X] \in V$ and $E[f(X)] \in \mathbb{R}$. Assume that $E[h(X)] = h(E[X])$ for every continuous linear functional $h:V\rightarrow\mathbb{R}$. Then
a) $E[X]\in \mathcal{X}$.
b) $f(E[X]) \leq E[f(X)]$.
Proof:
Proof of a) Suppose $E[X] \notin \mathcal{X}$ (we reach a contradiction). Then $\mathcal{X}$ is a closed and convex subset of $V$ and $E[X]$ is a point in $V$ that is not in $\mathcal{X}$. The (strict) Hahn-Banach separation theorem ensures existence of a nonzero continuous linear function $h:V\rightarrow \mathbb{R}$ and a value $c$ such that $h(E[X])> c \geq h(x)$ for all $x \in \mathcal{X}$. Since $X \in \mathcal{X}$ we get
$$ h(E[X]) > c \geq h(X) \quad \mbox{(surely)} $$
taking expectations and using linearity of expectation gives
$$ h(E[X]) > c \geq E[h(X)] = h(E[X]) $$
a contradiction. $\Box$
Proof of b) Define the normed vector space $\tilde{V} = V \times \mathbb{R}$. Define the set
$$\mathcal{A} = \{(x, y) : x \in \mathcal{X}, y \geq f(x)\}$$
It can be shown that (i) $\mathcal{A}$ is a closed and convex subset of $\tilde{V}$; (ii) $(X, f(X)) \in \mathcal{A}$; (iii) $(E[X], E[f(X)])\in \tilde{V}$. Applying the result of part (a) gives
$$ E[(X, f(X))] = (E[X], E[f(X)]) \in \mathcal{A} $$
and so $f(E[X])\leq E[f(X)]$ by the defining property of set $\mathcal{A}$. $\Box$
As a special case of the above claim, we define $V = \mathcal{X}$ as the Banach space of functions $g$ with $||g||\leq 1$ and note that $V$ is a closed and convex set and $||\cdot||:V\rightarrow\mathbb{R}$ is a continuous and convex function. So $||E[g]||\leq E[||g||]$.
