- #1

bobby2k

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Hi, can you please check if my proof is correct?

Assume that a sequence of continuous functions [itex]\{f_n\}[/itex] converges uniformly to [itex]f[/itex].

[itex]f_n: A\rightarrow B[/itex]. Also assume that [itex]A[/itex] is compact.

Then [itex]\{f_n\}[/itex] is equicontinuous.

Given an [itex]\epsilon[/itex] we must find a [itex]\delta[/itex] so that if [itex]x,y \in A,d_A(x,y)< \delta[/itex], then [itex]d_B(f_n(x),f_n(y))<\epsilon[/itex], for all [itex]n[/itex], and for all [itex]x,y[/itex].

Given [itex]\epsilon[/itex]. There exists an [itex]N[/itex] such that [itex]d_B(f_n(x),f(x))<\epsilon/3[/itex] for all [itex]n\ge N[/itex] and all [itex]x \in A[/itex].

Since f must be continuous and A is compact, f is uniformly continuous. So there exists a [itex]\delta^*[/itex] so that if [itex]d_A(x,y)<\delta^*\rightarrow d_B(f(x),f(y))<\epsilon/3[/itex]

For each [itex]i, i <N[/itex], there is a [itex]\delta_i[/itex], so that if [itex]d_A(x,y)<\delta_i \rightarrow d_B(f_i(x),f_i(y))<\epsilon[/itex]. This must be, because each [itex]f_n[/itex] must be uniformly continuous on the compact space A.

Choose [itex]\delta = min\{\delta^*,\delta_1,\delta_2,\delta_{N-1}\}[/itex].

Now, for any n, if [itex]d_A(x,y)<\delta \rightarrow d_B(f_n(x),f_n(y))< \epsilon[/itex]. This is so, because if [itex]n < N[/itex] it obviously holds, if [itex]n \ge N[/itex] we have:

[itex]d_B(f_n(x),f_n(y)) \le d_B(f_n(x),f(x))+d_B(f(x),f(y))+d_B(f(y),f_n(y))< \epsilon/3+\epsilon/3+\epsilon/3 = \epsilon[/itex].

**Proposition**Assume that a sequence of continuous functions [itex]\{f_n\}[/itex] converges uniformly to [itex]f[/itex].

[itex]f_n: A\rightarrow B[/itex]. Also assume that [itex]A[/itex] is compact.

Then [itex]\{f_n\}[/itex] is equicontinuous.

**Proof:**Given an [itex]\epsilon[/itex] we must find a [itex]\delta[/itex] so that if [itex]x,y \in A,d_A(x,y)< \delta[/itex], then [itex]d_B(f_n(x),f_n(y))<\epsilon[/itex], for all [itex]n[/itex], and for all [itex]x,y[/itex].

Given [itex]\epsilon[/itex]. There exists an [itex]N[/itex] such that [itex]d_B(f_n(x),f(x))<\epsilon/3[/itex] for all [itex]n\ge N[/itex] and all [itex]x \in A[/itex].

Since f must be continuous and A is compact, f is uniformly continuous. So there exists a [itex]\delta^*[/itex] so that if [itex]d_A(x,y)<\delta^*\rightarrow d_B(f(x),f(y))<\epsilon/3[/itex]

For each [itex]i, i <N[/itex], there is a [itex]\delta_i[/itex], so that if [itex]d_A(x,y)<\delta_i \rightarrow d_B(f_i(x),f_i(y))<\epsilon[/itex]. This must be, because each [itex]f_n[/itex] must be uniformly continuous on the compact space A.

Choose [itex]\delta = min\{\delta^*,\delta_1,\delta_2,\delta_{N-1}\}[/itex].

Now, for any n, if [itex]d_A(x,y)<\delta \rightarrow d_B(f_n(x),f_n(y))< \epsilon[/itex]. This is so, because if [itex]n < N[/itex] it obviously holds, if [itex]n \ge N[/itex] we have:

[itex]d_B(f_n(x),f_n(y)) \le d_B(f_n(x),f(x))+d_B(f(x),f(y))+d_B(f(y),f_n(y))< \epsilon/3+\epsilon/3+\epsilon/3 = \epsilon[/itex].

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