= Solution
The <Cameron-Martin space of a Gaussian random variable in a Banach space> is most conveniently defined using the <first Gaussian chaos>. Work over the real <separable Banach space> $B$, with continuous <dual space> $B^*$. For a centered <Gaussian random variable in a Banach space>, define
$$
\mathcal G=\overline{\{\ell(X):\ell\in B^*\}}^{L^2(\Omega)},\qquad Sg=\mathbb E[Xg]\quad(g\in\mathcal G).
$$
The set inside the closure is already a <linear subspace>. Every $g\in\mathcal G$ is a centered <Gaussian random variable>. The <Bochner integral> defining $S$ exists: <Fernique's theorem> states that for some $a>0$, $\mathbb E e^{a\|X\|_B^2}<\infty$, and then <Cauchy-Schwarz inequality> gives $\mathbb E\|Xg\|_B<\infty$.
The map $S$ is injective. Indeed, $Sg=0$ implies $\mathbb E[g\ell(X)]=\ell(Sg)=0$ for all $\ell\in B^*$. Density of these observations in $\mathcal G$ then implies $\mathbb Eg^2=0$. Define the <Reproducing kernel Hilbert space> by
$$
H=S\mathcal G\subset B,\qquad \langle Sg,Sv\rangle_H=\mathbb E[gv],\qquad \|Sg\|_H=\|g\|_2.
$$
The injectivity makes the <inner product> unambiguous, and the completeness of $\mathcal G$ makes $H$ a <Hilbert space>. The embedding into $B$ is continuous since
$$
\|Sg\|_B\leq(\mathbb E\|X\|_B^2)^{1/2}\|g\|_2.
$$
For $\ell\in B^*$ put $Q\ell=S(\ell(X))$. The <reproducing property> is
$$
\langle h,Q\ell\rangle_H=\ell(h),\qquad \langle Q\ell,Qm\rangle_H=\mathbb E[\ell(X)m(X)].
$$
This defines the <Banach-space Gaussian RKHS> even for a degenerate <Gaussian measure>.
For $h=Sg\in H$, write $\widehat h(X)=g$. There is a measurable coordinate $\widehat h$ on $B$, obtained as a limit in <L2 space> of continuous linear observations, and $\widehat h(X)$ has <normal distribution> $\mathcal N(0,\|h\|_H^2)$. The <Cameron-Martin theorem for a Gaussian measure>, in its real <separable Banach space> form, states that if $\mu$ is the <probability law> of $X$, then the <probability law> $\mu_h$ of $X+h$ is an <equivalent probability measure> to $\mu$ precisely when $h\in H$. For such a shift,
$$
\frac{d\mu_h}{d\mu}(x)=\exp\!\left(\widehat h(x)-\frac12\|h\|_H^2\right)\quad\text{for }\mu\text{-almost every }x.
$$
For $h\notin H$ the two laws are <mutually singular measures>. The coordinate $\widehat h$ need not be a continuous functional on $B$, and the sample $X$ need not belong to $H$.
Apply this theorem to the shift $-h$. With $a=\|h\|_H^2/2$, it gives
$$
\mathbb P(X-h\in C)=e^{-a}\mathbb E[\mathbf1_C(X)e^{-\widehat h(X)}].
$$
Choose the measurable coordinate so that $\widehat h(-x)=-\widehat h(x)$ on a symmetric set of full $\mu$-measure. This is possible by taking an almost-surely convergent subsequence of the approximating linear observations and intersecting its convergence set with its negative. Alternatively the joint law of $(X,\widehat h(X))$ is invariant under simultaneous negation, directly from those observations and their <mean-square convergence>. Since $C=-C$ and the centered <Gaussian measure> is symmetric, the two exponential integrals with signs $+$ and $-$ agree. Consequently
$$
\mathbb P(X-h\in C)=e^{-a}\mathbb E[\mathbf1_C(X)\cosh\widehat h(X)]\geq e^{-a}\mathbb P(X\in C),
$$
because $\cosh v\geq1$. The <symmetric Gaussian translation lower bound> is therefore
$$
\boxed{\mathbb P(X-h\in C)\geq e^{-\|h\|_H^2/2}\mathbb P(X\in C).}
$$
Only symmetry and Borel measurability of $C$ are required; <convexity> is not needed.
There is one degenerate flaw in the last printed claim. The <separable Banach space> $B=\{0\}$ with $X=0$ has $H=\{0\}$ dense in $B$, and all positive-radius balls have <probability> one. Nevertheless $\Phi(t)=1$ for every $t>0$. Thus the statement needs the additional hypothesis \b[$B\ne\{0\}$]. If the course convention excludes the zero space, this hypothesis is already implicit.
Under this necessary hypothesis, density of $H$ implies that $H$ contains a nonzero vector. Given $0<a<b$, scale that vector to obtain $h\in H$ with $\|h\|_B=(a+b)/2$, and choose $0<\delta<(b-a)/2$. The <triangle inequality> gives
$$
\|x-h\|_B\leq\delta\quad\Longrightarrow\quad a<\|x\|_B<b.
$$
The centered closed ball $C_\delta=\{x:\|x\|_B\leq\delta\}$ is a symmetric <Borel set>. Apply the <symmetric Gaussian translation lower bound> to get
$$
\begin{aligned}
\Phi(b)-\Phi(a)&=\mathbb P(a<\|X\|_B\leq b)\\
&\geq\mathbb P(\|X-h\|_B\leq\delta)\\
&\geq e^{-\|h\|_H^2/2}\Phi(\delta)>0.
\end{aligned}
$$
Hence \b[the Gaussian norm distribution is strictly increasing when $B$ is nonzero]. This proves the <strict increase of a Gaussian norm distribution> without assuming that its distribution has a density or that spheres have zero <probability>.
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