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Proof of Semicontinuity and Continuity Under Composition with a Continuous Map

lemmalem:semicontinuity-composition-continuous-2026a
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Reason: First published version. Direct epsilon-delta chaining of the two defining conditions; the restriction claim is the case of the inclusion map.

Proof

Fix yBy\in B and a real number ε\varepsilon with 0<ε0<\varepsilon; the point g(y)g(y) lies in AA. Since gg is continuous at yy relative to BB, for every real number η\eta with 0<η0<\eta there is a real number δ\delta with 0<δ0<\delta such that every yBy'\in B with dY(y,y)<δd_Y(y,y')<\delta satisfies dX(g(y),g(y))<ηd_X(g(y'),g(y))<\eta, hence also dX(g(y),g(y))<ηd_X(g(y),g(y'))<\eta by condition 3 in the definition of a metric. We use this repeatedly.

Claim 1. Suppose uu is upper semicontinuous at g(y)g(y) relative to AA. Choose η\eta with 0<η0<\eta such that every aAa\in A with dX(g(y),a)<ηd_X(g(y),a)<\eta satisfies u(a)<u(g(y))+εu(a)<u(g(y))+\varepsilon, and choose δ\delta for this η\eta as above. If yBy'\in B and dY(y,y)<δd_Y(y,y')<\delta, then g(y)Ag(y')\in A and dX(g(y),g(y))<ηd_X(g(y),g(y'))<\eta, so

(ug)(y)=u(g(y))<u(g(y))+ε=(ug)(y)+ε.(u\circ g)(y')=u(g(y'))<u(g(y))+\varepsilon=(u\circ g)(y)+\varepsilon .

Hence ugu\circ g is upper semicontinuous at yy relative to BB. As yBy\in B was arbitrary, ugu\circ g is upper semicontinuous on BB.

Claim 2. Suppose uu is lower semicontinuous at g(y)g(y) relative to AA. Choose η\eta with 0<η0<\eta such that every aAa\in A with dX(g(y),a)<ηd_X(g(y),a)<\eta satisfies u(g(y))ε<u(a)u(g(y))-\varepsilon<u(a), and choose δ\delta for this η\eta. If yBy'\in B and dY(y,y)<δd_Y(y,y')<\delta, then as before (ug)(y)ε<(ug)(y)(u\circ g)(y)-\varepsilon<(u\circ g)(y'). Hence ugu\circ g is lower semicontinuous at yy relative to BB, and therefore on BB.

Claim 3. Suppose uu is continuous at g(y)g(y) relative to AA. Choose η\eta with 0<η0<\eta such that every aAa\in A with dX(g(y),a)<ηd_X(g(y),a)<\eta satisfies dR(u(a),u(g(y)))<εd_{\mathbb{R}}(u(a),u(g(y)))<\varepsilon, and choose δ\delta for this η\eta. If yBy'\in B and dY(y,y)<δd_Y(y,y')<\delta, then dR((ug)(y),(ug)(y))<εd_{\mathbb{R}}((u\circ g)(y'),(u\circ g)(y))<\varepsilon. Hence ugu\circ g is continuous at yy relative to BB, and therefore on BB.

Claim 4. Apply claims 1, 2 and 3 with (Y,dY)(Y,d_Y) taken to be (X,dX)(X,d_X), with BB taken to be CC, and with g:CXg:C\to X the map g(c)=cg(c)=c. This map satisfies g(c)Ag(c)\in A for every cCc\in C because CAC\subseteq A, and it is continuous on CC relative to CC: given cCc\in C and a real number ε\varepsilon with 0<ε0<\varepsilon, take δ=ε\delta=\varepsilon, and note that cCc'\in C with dX(c,c)<δd_X(c,c')<\delta satisfies dX(g(c),g(c))=dX(c,c)=dX(c,c)<εd_X(g(c'),g(c))=d_X(c',c)=d_X(c,c')<\varepsilon by condition 3 in the definition of a metric. Since ug=uCu\circ g=u|_{C}, the three assertions of claim 4 follow.

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