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Janet Forbes

Janet Forbes

Answered question

2022-07-14

Let f : [ a , b ] R be a continuous function such that for each x [ a , b ] there exists y [ a , b ] such that | f ( y ) | 1 2 | f ( x ) | . Prove that there exists c [ a , b ] such that f ( c ) = 0.

I am trying to use intermediate value theorem for continuous function. Here | f ( x ) | is maximum with respect to | f ( y ) | . Hence
| f ( y ) | 1 / 2 ( | f ( x ) | + | f ( y ) | ) | f ( x ) | .
Now I can say f ( c ) = 1 / 2 ( | f ( x ) | + | f ( y ) | for some c ( a , b ) . But how to show f ( c ) = 0 ?

Answer & Explanation

Jayvion Mclaughlin

Jayvion Mclaughlin

Beginner2022-07-15Added 14 answers

Let x 0 [ a , b ]. Then there is x 1 [ a , b ] such that | f ( x 1 ) | 1 2 | f ( x 0 ) | .

Now we get inductively a sequence ( x n ) in [ a , b ] with
(*) | f ( x n ) | 1 2 n | f ( x 0 ) | .
( x n ) contains a convergent subsequence ( x n k ) with limit c [ a , b ].

f is continuous, thus f ( x n k ) f ( c ) .

From (*) we get: f ( x n k ) 0.

Consequence: f ( c ) = 0

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