Coercivness of a function involving $x$ and $\|x\|$ This problem often arises in machine learning. I'm looking at a regularized optimization problem of the form
$$T(f) = L(f) + \lambda \|f\|^2 \tag{1}$$
where $f$ belongs to some reproducing kernel Hilber space. $L$ is some type of loss function. I'm interested in $L(f)=\sum_i^I (y - b_if(x_i))^2$ where $(x_i,y_i)$ are  input / output pairs and are fixed. Note we assume $\lambda>0$. If one want to minimise $(1)$ often one uses the following theorem

Theorem If $T$ is coercive, lower semi-continuous and convex then a global minimizer exists.

It is clear that $T$ is convex in $f$. For lower semi-continuity is it correct that we need the fact that $f$ belongs to a RKHS. Because then we have for any $f_n\to f$ as well $f_n(x) \to f(x),\forall x$. Or how can one prove lower semi-continuity?
The last question is about coercivness of $T$. Clearly, $\lambda \|f\|^2\to\infty$ for $\|f\|\to\infty$. But the part which is bothering me is. How can I guarantee that $\sum_i^I (y - b_if(x_i))^2$ does not affect this? I'm not fully clear what the definition of coercivness means for point values $f(x_i)$.
 A: If $t$ is bounded from below by an affine function, i.e., there is $f\in H$, $a\in \mathbb R$ such that
$$
t(x) \ge \langle f,x\rangle + a \quad \forall x
$$
then $T$ is coercive. A sufficient condition for this boundedness is: $t$ is lower semicontinuous and convex.
A: Coerciveness follows from the fact that $L(f)$ is non-negative. So for any sequence $f_n$ such that $\Vert f_n\Vert \to \infty$,
$$\lim_{n} T(f_n) \ge \lim_n\lambda \Vert f_n \Vert^2 = \infty.$$
Continuity can be shown by using properties of RKHS, assuming that the number of points in $I$ is finite. Let $f_n$ be a sequence that converges to $f$ in norm. Convergence in RKHS implies pointwise convergence, $f_n(x)\to f(x)$ for all $x$ (see Cor. 21 here), so $(y-b_i f_n(x_i))^2 \to(y-b_i f(x_i))^2$ for every $i\in I$. Since $L$ is a finite sum of such functions, and a finite sum of continuous functions is continuous, $f \mapsto L(f)$ is continuous.
It is also the case that $\Vert f_n\Vert \to \Vert f\Vert$, which follows from the inequalities
\begin{align}\Vert f \Vert - \Vert f_n - f \Vert \le \Vert f_n \Vert \le \Vert f \Vert + \Vert f_n - f \Vert
\end{align}
and $\Vert f_n - f\Vert \to 0$ by definition of convergence in Hilbert space. Thus, $f \mapsto \lambda \Vert f\Vert^2$ is continuous. Since the sum of two continuous functions is continuous, $f \mapsto T(f)$ is continuous.
