Uniform convergence with Lp functions

Question

I have a convergence question:

Say we have a sequence of functions $\{u_{m}\}_{m=1}^{\infty}$ where $\{u_{m}\}_{1}^{\infty} \subset L^{p}(U)$ and where $U$ is bounded. Consider $u^{\epsilon} := \eta_{\epsilon}\ast u_{m}$ where $\eta_{\epsilon}$ is the usual mollifier.

If I was able to show that for every $m$ we have $||u_{m}^{\epsilon}-u_{m}||_{L^{1}(U)} \leq \epsilon C$ for some constant $C$, then clearly I have shown that $u_{m}^{\epsilon} \rightarrow u_{m}$ in $L^{1}(U)$ as $\epsilon \rightarrow 0$. My question is simply how is the convergence "uniform in m"?

My understanding of uniform convergence is using the definition:

If $f_{n}: S \rightarrow \mathbb{R}$ converges uniformly to $f: S \rightarrow \mathbb{R}$ iff $\alpha_{n} = \text{sup}_{x}|f_{n}(x)-f(x)|$ tends to 0.

I don't see how this definition coincides with the above.

Thanks!

Answer 1

The convergence being "uniform in $m$" here means that given any $\delta > 0$, you can find an $\varepsilon_0 > 0$, such that for all $0 < \varepsilon \leqslant \varepsilon_0$, you have

$$\lVert u_m^\varepsilon - u_m\rVert_{L^1(U)} < \delta$$

for all $m$.

Or,

$$\sup_{m} \lVert u_m^\varepsilon - u_m\rVert_{L^1(U)} \xrightarrow{\varepsilon\to 0} 0.$$