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Let $f$ be a finite convex function on a neighbourhood of $x\in {\mathbb{R}}^d$, and suppose its first partial derivatives exist at $x$. For any fixed vector $h$, convexity givesThe equality uses the one-dimensional first-order Taylor expansion along each coordinate line; no continuity of second partial derivatives is assumed. Applying this upper bound to $-h$ and using $f(x)\le [f(x+th)+f(x-th)]/2$ gives the matching lower bound. Hence the displayed directional expansion holds. This supplies the expansion used in the supporting inequality for a differentiable convex function even when continuous second partial derivatives have not been postulated.

Let $m=\mathbb{E}X$. First, $m$ lies in the open interval $I$. For example, if $I$ has a finite lower endpoint $\ell$, then $X-\ell >0$ almost surely, and its expected value is strictly positive; hence $m>\ell$. The corresponding argument at a finite upper endpoint gives the other strict inequality. Integrability ensures $m$ is finite when either endpoint is infinite.

The tangent affine function at $m$ gives the supporting inequality for a differentiable convex function. Under the stipulated condition $f^{\prime \prime }>0$, contact with that tangent occurs only at $m$: defineThen $h(m)=0$, and $h(t)>0$ for every $t\ne m$. One direct proof uses $h^{\prime }(t)=f^{\prime }(t)-f^{\prime }(m)$: $f^{\prime }$ is strictly increasing, so $h$ decreases towards $m$ on the left and increases away from $m$ on the right. Thereforewith equality exactly when $h(X)=0$ almost surely, equivalently when $X=m$ almost surely. Conversely, an almost surely constant $X=m$ plainly gives equality. This proves the strict Jensen equality criterion.

No extra assumption that $f(X)$ is an integrable random variable is needed to interpret this argument. Its negative part is bounded by the absolute value of the integrable random variable $f(m)+f^{\prime }(m)(X-m)$, since $h(X)\ge 0$. Thus its expected value is defined in $(-\infty ,+\infty ]$; if it equals the finite number $f(m)$, then $\mathbb{E}[h(X)]=0$ and the required integrability follows automatically. Equality holds if and only if $X=\mathbb{E}X$ almost surely.

Put $\mathbf{h}=\mathbf{y}-\mathbf{x}$. For $0<t\le 1$, convexity gives $f(\mathbf{x}+t\mathbf{h})\le (1-t)f(\mathbf{x})+tf(\mathbf{y})$, henceBy the first-order form of Taylor's theorem, the right side tends to ${\mathbf{h}}^{\mathsf{T}}\nabla f(\mathbf{x})$ as $t\downarrow 0$. ThereforeThis is the supporting inequality for a differentiable convex function. The stated partial derivatives are enough: coordinate derivatives give directional Taylor expansion for convex functions, so no unmentioned continuity assumption on the second partial derivatives is needed. The result in fact needs less smoothness than the question assumes.