Continuity of one function, implies continuity of another?

  • Thread starter bobby2k
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  • #1
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Hi

Lets say that f(x) is continuous. Then [itex] \int_0^x \! f(t)dt=G(x)[/itex] is continuous. (I don't think you have to say that f need to be continuous for this, all we need to say is that f is integrable?, or do we need continuity of f here?)

But my main question is about the converse. lets say that [itex] \int_0^x \! f(t)dt=G(x)[/itex] is continuous, does that imply that f is continuous?

Have a nice sunday.
 
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  • #2
pasmith
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Hi

Lets say that f(x) is continuous. Then [itex] \int_0^x \! f(t)dt=G(x)[/itex] is continuous. (I don't think you have to say that f need to be continuous for this, all we need to say is that f is integrable?, or do we need continuity of f here?)

This is true for any integrable function.

But my main question is about the converse. lets say that [itex] \int_0^x \! f(t)dt=G(x)[/itex] is continuous, does that imply that f is continuous?

No: by the above, [itex]f[/itex] does not need to be continuous for its integral to be continuous.
 
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  • #3
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This is true for any integrable function.



No: by the above, [itex]f[/itex] does not need to be continuous for its integral to be continuous.

Hehe, ofcourse, thanks.
 

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