How to show function addition is continuous in metric space?

Question

We have the map $P(f,g)=f+g$, $C[0,1]^2 \rightarrow C[0,1]$, now I want to prove that it is continuous on $L^1$ metric.

The outline is find $\delta$, such that $d_1(d_1(f_1,g_1),d_1(f_2,g_2))<\delta$ implies $d_1(P(f_1,g_1),P(f_2,g_2))<\epsilon$.

I got $\left|\int_0^1|(f_1(t)-f_2(t))|\,\mathrm{d}t-\int_0^1|(g_1(t)-g_2(t))|\,\mathrm{d}t\right|$ for the left expression, and $\int_0^1|(f_1(t)-f_2(t))+(g_1(t)-g_2(t))|\,\mathrm{d}t$ for the right expression. However, using the triangle inequality, the inequality is opposite from the desired direction. Where is the mistake? It is because the definition for L1 is incorrect, or a different technique should be applied?

Answer 1

I reckon we equip $X =: C^0([0,1])$ with the norm $\Vert \,f\, \Vert = \int_{[0,1]} |\,f(s)|\,ds$ and $X^2$ with its natural product topology induced by $\Vert (f,g) \Vert_{\times} = \max\big\{\|f\|,\|g\|\big\}$. It suffices to prove sequential continuity. Let $(f_n,g_n) \rightarrow (f_0,g_0) \in X^2$, and we want to show that $P(f_n,g_n) \rightarrow P(f_0,g_0)$. Indeed:

\begin{align}\Vert P(f_0, g_0) - P(f_n,g_n) \Vert &= \int_{[0,1]}|f_n(s) - f_0(s) + g_n(s) - g_0(s)|ds \\&\leq \int_{[0,1]} |f_n(s) - f_0(s)| + |g_n(s) - g_0(s)|ds \\ &\leq \int_{[0,1]} |f_n(s) - f_0(s)|ds + \int_{[0,1]}|g_n(s) - g_0(s)|ds \end{align}

Since $\Vert (f_n - f_0,g_n-g_0) \Vert_{\times} \rightarrow 0$, the right-hand side goes to $0, finishing the proof.

Answer 2

First, you should be looking for $$ \lVert (f_1,g_1) - (f_2,g_2) \rVert_{L^1\times L^1} < \delta $$ not $$ d_1(d_1(f_1,g_1),d_1(f_2,g_2))<\delta. $$

Second, there is a sign error between the to integrals in your formula for the "left expression". If you fix that, the triangle equality will give you exactly what you need.