# Thread: is this sequence convergent?

1. ## is this sequence convergent?

Let Is the sequence convergent in ?
First I tried to check if this sequence is Cauchy and I calculated integral of the sequence over . I got ). Then I'm stuck can anyone help?

2. Originally Posted by hermanni

Let Is the sequence convergent in ?
First I tried to check if this sequence is Cauchy and I calculated integral of the sequence over . I got ). Then I'm stuck can anyone help?
If the $L_1$-limit of the sequence $f_n(t) = \frac1{\sqrt n}e^{-\frac1n(t-n)^2}$ exists, then it must be equal to the pointwise limit, which is the zero function. So all you need to do is to check whether $\|f_n\|_1 = \displaystyle\int_0^\infty \!\! f_n(t)\,dt \to0$ as $n\to\infty$. In fact, if you make the substitution $u = \frac1{\sqrt n}(t-n)$, you see that $\|f_n\|_1 = \displaystyle\int_{-\sqrt n}^\infty e^{-u^2}du \to\sqrt\pi$, and so the sequence does not converge in the $L_1$-norm.

3. What about the same $x_{n}$ , is ${ e^n x_{n} }$ is convergent in C[0,1] with sup norm? I think it converges to $e^{2t}$.

4. Originally Posted by hermanni
What about the same $x_{n}$ , is ${ e^n x_{n} }$ is convergent in C[0,1] with sup norm? I think it converges to $e^{2t}$.
I agree, the pointwise limit is $e^{2t}$. To show that this is the limit in the sup norm, you need to show that $e^n x_{n} \to e^{2t}$ uniformly on [0,1]. So look at the difference $e^n x_{n} - e^{2t}$ and see if that can be made small for all $t\in[0,1]$, provided that $n$ is large enough.

5. One more thing : can you advise good functional analysis books ? (especially on banach spaces )