Suppose that $H$ is a Hilbert space with norm $\|\cdot\|_H$, and $K : \mathbb R^n \times H \to \mathbb R$ is a function such that $K(x,y)$ is continuous in $x$ and a bounded linear operator in $y$. Need $f(x) = \|K(x,\cdot)\|$ be a continuous function, where $$\|K(x,\cdot)\| = \sup_{\|y\|_H = 1}|K(x,y)|?$$ If the family of functions $\mathcal F := \{K(\cdot,y) : y\in H\}$ is equicontinuous, then the answer is affirmative, since $$|f(b) - f(a)| \leq \|K(b,\cdot)-K(a,\cdot)\| \leq \varepsilon$$ for any $\varepsilon > 0$, as soon as $a$ is in a neighborhood of $b$ small enough that $|K(b,y) - K(a,y)| < \varepsilon$ for all $y\in H$. When $H = \mathbb R^m$ equipped with the Euclidean norm, then this assertion should be true as well. In this case, there exists a matrix-valued function $A : \mathbb R^n \to \mathbb R^{1\times m}$ such that $K(x,y) = A(x)y$ for every $(x,y)\in \mathbb R^{n+m}$. If we write $A(x) = \left[\alpha_1(x), \cdots, \alpha_m(x)\right]$ then we see $f(x) = \|A(x)\| = \sqrt{\alpha_1(x)^2 + \cdots + \alpha_m(x)^2}$, and since each $\alpha_i(x) := K(x,e_i)$ (where $e_i$ is a basis element of $\mathbb R^m$) is continuous, we see $f$ is continuous.
However, I have novice experience with Hilbert spaces, and I have some doubt that the assertion holds in a completely general context. Therefore I was hoping someone could point me in the correct direction of a proof, or help furnish a counter example.
The answer is no unfortunately. It is easier to work with sequences in this context hence we will need this lemma to built a counterexample,
Lemma : If $(u_n)_{n \geqslant 1}$ is a sequence of elements of $H^*$ that weakly converges toward $0$, there exists a continuous $K : \mathbb{R} \times H \rightarrow \mathbb{R}$ that is linear in the second variable such that for all $n \geqslant 1$, $K\!\left(\frac{1}{n},\cdot\right) = u_n$ and $K(0,\cdot) = 0$.
Indeed, let us define for all $x \geqslant 1$, $K(x,y) = u_1(y)$, for all $x \leqslant 0$, $K(x,y) = 0$ and for all $0 < x \leqslant 1$, let $n = \left\lfloor\frac{1}{x}\right\rfloor$ is the only integer that verifies $\frac{1}{n + 1} < x \leqslant \frac{1}{n}$ and, $$ K(x,y) = \frac{x - (n + 1)^{-1}}{n^{-1} - (n + 1)^{-1}}u_n(y) + \frac{n^{-1} - x}{n^{-1} - (n + 1)^{-1}}u_{n + 1}(y). $$ $K$ is clearly linear in the second variable and each $K(x,\cdot)$ is continuous. Moreover, $K(\cdot,y)$ is continuous for each fixed $y$ because it is affine by part (the continuity at $x = 0$ is ensured by the weak convergence of $u_n$ toward $0$). The sequence $(u_n)$ is uniformly bounded by the uniform boundedness principle hence for all $0 < x \leqslant 1$, \begin{align*} \|K(x,\cdot)\| & = \left\|\frac{x - (n + 1)^{-1}}{n^{-1} - (n + 1)^{-1}}u_n + \frac{n^{-1} - x}{n^{-1} - (n + 1)^{-1}}u_{n + 1}\right\|\\ & \leqslant \left|\frac{x - (n + 1)^{-1}}{n^{-1} - (n + 1)^{-1}}\right|\|u_n\| + \left|\frac{n^{-1} - x}{n^{-1} - (n + 1)^{-1}}\right|\|u_{n + 1}\|\\ & \leqslant \frac{x - (n + 1)^{-1}}{n^{-1} - (n + 1)^{-1}}\sup_{n}\|u_n\| + \frac{n^{-1} - x}{n^{-1} - (n + 1)^{-1}}\sup_n\|u_n\|\\ & = \sup_n\|u_n\|, \end{align*} and the same equality trivially holds for $x \leqslant 0$ and $x \geqslant 1$. It implies the total continuity of $K$ because for all $x,x',y,y'$, \begin{align*} |K(x',y') - K(x,y)| & \leqslant |K(x',y') - K(x',y)| + |K(x',y) - K(x,y)|\\ & \leqslant \sup_n\|u_n\|\|y' - y\| + |K(x',y) - K(x,y)|\\ & \rightarrow 0, \end{align*} when $(x',y') \rightarrow (x,y)$ by continuity of $K$ with respect to $x$ when $y$ is fixed. It proves the lemma.
Now, to build the counterexample, consider for example $H = \mathcal{l}^2(\mathbb{N})$ and for all $n$, $u_n(y) = y_n$, which weakly converges toward $0$ but it doesn't converge strongly because for all $n$, $\|u_n\| = 1$. Let $K$ be given by the lemma. We have for all $n \geqslant 1$, $\left\|K\!\left(\frac{1}{n},\cdot\right)\right\| = \|u_n\| = 1$ but $\|K(0,\cdot)\| = 0$.
Notice however that such a $K : \mathbb{R}^n \times H \rightarrow \mathbb{R}$ is always lower semi-continuous. Indeed, if $x_m \rightarrow x$ in $\mathbb{R}^n$, then for all $y$, $K(x_m,y) \rightarrow K(x,y)$ hence $K(x_m,\cdot) \rightarrow K(x,\cdot)$ weakly so $\|K(x,\cdot)\| \leqslant \liminf_m\|K(x_m,\cdot)\|$ by weak convergence.