Using Weierstrass theorem to prove a function has at least one optimal soluntion

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Problem: Given a matrix $X \in \mathbb{R}^{m \times n}$ and a function $F: \mathbb{R}^{m\times n} \to \mathbb{R}$ defined as following $$F(X) = \dfrac{\sum_{i=t+1}^m \vert \sigma_i(X)\vert}{\sqrt{\sum_{i=t+1}^m \sigma_i^2(X)} } + \Vert \mathcal{A}(X)-b\Vert_2^2,$$ where $\sigma_i(X)$ are singular values of $X$, $b \in \mathbb{R}^p$, $\mathcal{A}:\mathbb{R}^{m\times n} \to \mathbb{R}^p$. Prove that $F$ has at least one optimal solution.

My attempt:

Case 1: $t \ge m-1$. Thus, the objective value is $$F(X) = 1 + \Vert \mathcal{A}(X)-b\Vert_2^2.$$ We can easily see that $X = \mathcal{A}^{\dagger}(b)$ denote as Moore–Penrose pseudoinverse of $\mathcal{A}$.

Case 2: $t <m-1$. I am going to use the Weierstrass's theorem to prove $F$ has as least one optimal soluntion.

  1. Firstly, $F$ is proper.
  2. Secondly, $F$ is continuous.
  3. Thirdly, I try to prove that $F$ is coercive. Let $\left\{X^k\right\}$ is a sequence such that $F(X^k)$ is bounded. So we need to prove that $\left\{X^k\right\}$ is bounded. This is the part I still can't solve. I don't know where to start.

Hope you can advise me on some way to break through this.