L^2 integral problem.

**younhock** · December 15th 2011, 05:14 AM

Suppose that $fg \in L^2 ([a,b])$ for all $f \in L^2([a,b])$ . How to show that $g$ is also in $L^2([a,b])$

**emakarov** · December 15th 2011, 05:18 AM

Take f(x) = 1.

**younhock** · December 15th 2011, 05:21 AM

Originally Posted by emakarov

Take f(x) = 1.

this doesn't show for every f.

**emakarov** · December 15th 2011, 05:28 AM

Originally Posted by younhock

Suppose that $fg \in L^2 ([a,b])$ for all $f \in L^2([a,b])$ . How to show that $g$ is also in $L^2([a,b])$

I thought you assume that $fg \in L^2 ([a,b])$ for all $f \in L^2([a,b])$ . We have $1 \in L^2([a,b])$ .

**younhock** · December 15th 2011, 05:31 AM

Originally Posted by emakarov

I thought you assume that $fg \in L^2 ([a,b])$ for all $f \in L^2([a,b])$ . We have $1 \in L^2([a,b])$ .

Sorry I not really get what you mean. Would you explain in details?

**emakarov** · December 15th 2011, 05:46 AM

You are given that $fg \in L^2 ([a,b])$ for all $f \in L^2 ([a,b])$ . This is known; you don't have to prove this. Therefore, you can replace $f$ with any function, as long as it is in $L^2 ([a,b])$ , and still get a true statement. I suggest replacing $f$ with a function that equals 1 everywhere.

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