# Thread: Limit of a Sequence of Elements in L^2

1. ## Limit of a Sequence of Elements in L^2

I am asked to prove: If $\displaystyle$\displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$$then \displaystyle \displaystyle \lim_{n \rightarrow \infty} ||f_{n}|| = ||f||$$. I've been trying to answer this by an epsilon-delta argument, but I wonder if this is the best way. Should I try to do some algebra, converting $\displaystyle ||f_{n} - f||$ into an integral and then try to push the limit inside? If so, I'm not certain what permits me to move the limit inside a root.

2. Progress on the epsilon-delta argument:

$\displaystyle$\displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = \lim_{n \rightarrow \infty}\Bigl( \int_{I} f_{n}^{2}(x) - 2f_{n}(x)f(x) + f^{2}(x) \Bigr)^{\frac{1}{2}}$.$ Let $\displaystyle$\varepsilon > 0$$be given, then there exists an \displaystyle N \in \mathbb{N}$$ such that $\displaystyle$\displaystyle n \geq N \Rightarrow \Bigl( \int_{I} f_{n}^{2}(x) - 2f_{n}(x)f(x) + f^{2}(x) \Bigr)^{\frac{1}{2}} < \varepsilon \Rightarrow \int_{I} f^{2}_{n}(s) - 2\int_{I}f_{n}(x)f(x) + \int_{I} f^{2}(x) < \varepsilon^{2}$$. On the other hand, \displaystyle \displaystyle \lim_{n \rightarrow \infty} ||f_{n}|| = ||f|| \Leftarrow ||f_{n}|| - ||f|| = \Bigl| \Bigl( \int_{I} f_{n}^{2} \Bigr)^{\frac{1}{2}} - \Bigl( \int_{I} f^{2} \Bigr)^{\frac{1}{2}} \Bigr| < \varepsilon \Leftarrow \int_{I} f^{2}(x) - 2 \Bigl( \int_{I} f_{n}(x)f(x) \Bigr)^{\frac{1}{2}} + \int_{I} f^{2} < \varepsilon^{2}$$. It seems that all I have to do is deal with the square root, but since the quantities may be greater or less than 1, I'm not sure how to proceed.

3. $\displaystyle ||\cdot||$ is a norm, so you can use the triangle inequality :
$\displaystyle ||f_{n}-f||\geq|||f_{n}||-||f|||$

4. I think it's like this:

$\displaystyle$\displaystyle \lim_{n \rightarrow \infty} ||f_{n}|| = ||f|| \Leftarrow ||f_{n}|| - ||f|| = \Bigl| \Bigl( \int_{I} f_{n}^{2} \Bigr)^{\frac{1}{2}} - \Bigl( \int_{I} f^{2} \Bigr)^{\frac{1}{2}} \Bigr| < \varepsilon \Leftarrow \int_{I} f^{2}(x) - 2 \Bigl( \int_{I} f_{n}(x)^2f(x)^2 \Bigr)^{\frac{1}{2}} + \int_{I} f^{2}(x) < \varepsilon^{2}$$From the first norm you have \displaystyle \displaystyle\int_{I} f^{2}_{n}(s) - 2\int_{I}f_{n}(x)f(x) + \int_{I} f^{2}(x) < \varepsilon^{2}$$

$\displaystyle$\displaystyle \Rightarrow \int_{I} f^{2}_{n}(s) + \int_{I} f^{2}(x) < \varepsilon^{2} + 2\int_{I}f_{n}(x)f(x)$$Substituting this into the top thing we get: \displaystyle \displaystyle \int_{I} f^{2}(x) - 2 \Bigl( \int_{I} f_{n}(x)^2f(x)^2 \Bigr)^{\frac{1}{2}} + \int_{I} f^{2}(x) < \varepsilon^{2} - 2 \Bigl( \int_{I} f_{n}(x)^2f(x)^2 \Bigr)^{\frac{1}{2}} + 2\int_{I}f_{n}(x)f(x)$$.

We gotta show $\displaystyle$\displaystyle- 2 \Bigl( \int_{I} f_{n}(x)^2f(x)^2 \Bigr)^{\frac{1}{2}} + 2\int_{I}f_{n}(x)f(x) < 0$$? I think it'll go along the lines of: \displaystyle \displaystyle \int_{I}f_{n}(x)f(x) < \int_{I}|f_{n}(x)f(x)| = \int_{I}(f_{n}^2(x)f(x)^2)^{1/2} < \Bigl( \int_{I} f_{n}(x)^2f(x)^2 \Bigr)^{\frac{1}{2}}$$

However, if you use what iondor posted, it becomes much easier.