Uniform Boundedness: Am I right or my TA?

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I am a student, and I disagree with the solutions our TA has prepared. I am seeking verification that I am correct or explanation as to why I am wrong. It seems to be a disagreement or misunderstanding of the definition of uniform boundedness.

The question boils down to this: For what choices of $a, b \in \mathbb{R}$, is the following risk function uniformly bounded as a function of $\mu$ (here, $\sigma$ is a constant)?

$$ R(\mu ; a, b) = a^2 + 2a(b-1)\mu + (b-1)^2\mu^2 + b^2 \frac{\sigma^2}{n}$$

Clearly, we must have $b=1$ to knock out the polynomial and linear terms in $\mu$. Once $b=1$, we have:

$$ R(\mu ; a, b=1) = a^2 + \frac{\sigma^2}{n} $$

Our TA says this is uniformly bounded for all choices of $a$. By my understanding, UNIFORM boundedness means there must exist one single bound $M > 0$ such that $ \mid R(\mu ; a, b=1) \mid \le M, \forall a, \forall \mu$. That is, our choice of the bound $M$ must be good for any choice of the parameter $a$.

Clearly there is no one bound to contain this function for all $a \in \mathbb{R} $: If we choose $M = 100$, choose $a=101$ (the other term is strictly positive), and our function is not bounded by M. I say we need $a \in [-1,1]$, so that $a^2 \le 1$, and then we can bound the family of functions with a constant $1+ \frac{\sigma^2}{n}$. It seems like he is thinking only of boundedness, not uniform boundedness.

Apologies for the trivial matter; but where else to go for a dispute over definitions than to the community? Thank you very much!

PS: the actual context is decision theoretic statistics and risk.