Let \(f_n:[0,1]\to\mathbb R\) be continuous and suppose \[ \int_0^1\bigl(f_n(x)-f_m(x)\bigr)^2\,dx\to0 \qquad(n,m\to\infty). \] Let \(K:[0,1]\times[0,1]\to\mathbb R\) be continuous, and define \[ g_n(x)=\int_0^1K(x,y)f_n(y)\,dy. \] Prove that \(\{g_n\}\) converges uniformly on \([0,1]\).
Solution
1. There is a constant \(M\geq0\) such that\[ \abs{K(x,y)} \leq M \]for every \((x,y)\in[0,1]^2\).
The kernel \(K\) is continuous on the compact set \([0,1]^2\). Hence the continuous function \[ (x,y)\longmapsto\abs{K(x,y)} \] attains a finite maximum. Let that maximum be \(M\).
2. For every \(m,n\) and every \(x\in[0,1]\),\[ \abs{g_n(x)-g_m(x)} \leq M \left( \int_0^1 \bigl(f_n(y)-f_m(y)\bigr)^2 \,dy \right)^{1/2}. \]
By the definition of \(g_n\), \[ \begin{aligned} \abs{g_n(x)-g_m(x)} &= \abs{ \int_0^1 K(x,y) \bigl(f_n(y)-f_m(y)\bigr) \,dy }\\ &\leq \left( \int_0^1 \abs{K(x,y)}^2 \,dy \right)^{1/2} \left( \int_0^1 \bigl(f_n(y)-f_m(y)\bigr)^2 \,dy \right)^{1/2} \end{aligned} \] by Cauchy–Schwarz. Step 1 gives \[ \int_0^1 \abs{K(x,y)}^2 \,dy \leq \int_0^1M^2\,dy = M^2, \] which yields the stated estimate.
3. One has\[ \sup_{x\in[0,1]} \abs{g_n(x)-g_m(x)} \longrightarrow 0 \]as \(m,n\to\infty\).
The right-hand side of the estimate in step 2 is independent of \(x\). Therefore \[ \sup_{x\in[0,1]} \abs{g_n(x)-g_m(x)} \leq M \left( \int_0^1 \bigl(f_n(y)-f_m(y)\bigr)^2 \,dy \right)^{1/2}. \] By hypothesis, the integral tends to zero as \(m,n\to\infty\), so the displayed supremum does also.
4. For every \(x\in[0,1]\), the sequence\[ \bigl(g_n(x)\bigr)_{n\geq1} \]converges in \(\RR\).
Step 3 shows in particular that, for fixed \(x\), \[ \abs{g_n(x)-g_m(x)} \longrightarrow 0 \] as \(m,n\to\infty\). Thus \(\bigl(g_n(x)\bigr)\) is a Cauchy sequence in the complete metric space \(\RR\), so it converges.
5. Define\[ g(x) = \lim_{n\to\infty}g_n(x) \]for \(x\in[0,1]\). Then\[ \boxed{ g_n\longrightarrow g \text{ uniformly on }[0,1]. } \]
Let \(\varepsilon>0\). By step 3, there is \(N\) such that \[ \abs{g_n(x)-g_m(x)} < \varepsilon \] for every \(x\in[0,1]\) whenever \(m,n\geq N\).
Fix \(n\geq N\) and \(x\in[0,1]\). Letting \(m\to\infty\) and using the definition of \(g(x)\) from step 4 gives \[ \abs{g_n(x)-g(x)} \leq \varepsilon. \] This holds for every \(x\in[0,1]\), so \[ \sup_{x\in[0,1]} \abs{g_n(x)-g(x)} \leq \varepsilon \] whenever \(n\geq N\). This is uniform convergence.
6. Q.E.D.
Step 5 is the required conclusion.