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For any continuous function $f(x)$ we have that $$ (*) \quad \quad \lim_{x\to x_0} f(x) = f(\lim_{x\to x_0}x) = f(x_0). $$ Because of this, I always believed that by the continuity of the norm function, for a sequence of functions $f_n$ converging to some function $f$, that we could say $$ (**) \quad \quad \lim_{n\to \infty} ||f_n||_p = || \lim_{n\to \infty} f_n||_p = ||f||_p. $$ So is there a difference between $(*)$ and $(**)$? Are we not allowed to perform the switching of limits in $(**)$, even though the norm is continuous?

On a related note I also always thought we could do the following for a sequence of functions in a Hilbert space: $$ \lim_{n\to \infty} \langle f_n, g \rangle = \langle \lim_{n\to \infty} f_n, g \rangle = \langle f, g \rangle. $$ Is this also not true?

ManUtdBloke
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  • In order to justify (**), you would need that $f_n \stackrel{L^p}{\longrightarrow} f$, not pointwise. Similarly, for the last question about Hilbert spaces, it is true if $f_n \to f$ in the Hilbert space, but not necessarily if $f_n \to f$ pointwise. – shalop Sep 01 '17 at 17:05
  • In this post - https://math.stackexchange.com/questions/133678/showing-that-ker-t-is-closed-if-and-only-if-t-is-continuous - the poster uses exactly the property I asked about. For a linear operator $T:X\to \mathbb{R}$, and a sequence $x_n\to x$, he says that 'by continuity' $$T(x) = \lim_{n\to \infty} T(x_n) = \dots$$ The people who answered the question said that what he did was fine. So this user didn't have to specify any constraints on the operator $T$. So what is the difference between his situation and my question..why are there constraints in my case and not his? – ManUtdBloke Sep 02 '17 at 19:00
  • In the linked Q it states that if $T$ is continuous then ..., which is a constraint. – DanielWainfleet Sep 03 '17 at 16:42
  • @DanielWainfleet I should have said additional constraints. The operator $T$ in that question is only constrained to be continuous, and he is then allowed to 'bring the limit inside'. In my case I also have a continuous operator, i.e. the norm, and going by the answers to that question, that would be sufficient for me to bring the limit inside. So are the answers to that question incorrect? Why can he bring the limit inside the continuous operator $T$, yet I can't bring the limit inside the continuous operator $||\cdot||$? – ManUtdBloke Sep 06 '17 at 06:55
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    If $\lim_{n\to \infty}|f_n-f|=0$ (which is uniform convergence, or convergence in norm) then $\lim_{n\to \infty} |f_n|=|f|.$..However if $\lim_{n\to \infty}f_n(x)=f(x)$ for each $x$ (point-wise convergence) it may be that $|f_n|$ does not converge to $|f|.$ E.G. if ${e_n:n\in N}$ is an ortho-normal Hibert-space basis and $f_n(x)=<x,e_n>$ (inner product) then $f_n(x)\to 0$ for each $x$ but $|f_n|=1$ for all $n$.... There are different kinds of convergence of a sequence of functions. – DanielWainfleet Sep 06 '17 at 08:18

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In the setting of functional analysis, one needs to understand the notation $$ \lim_{n\to\infty}f_n=f\tag{1} $$ with contexts since there are different modes of convergence. Usually, (1) is understood as norm convergence in some Banach space. For other modes of convergence, one would write $$ f_n\to f\quad\hbox{ [in some mode]} $$ where [in some mode] could be "almost everywhere", "pointwise", "in measure", "weakly", etc. assuming one has extra appropriate structures.

The statement ($**$) in your question should be interpreted precisely as

if $f_n\to f$ in norm $\|\cdot\|_p$, which by definition means $\|f_n-f\|_p\to 0$, then $\|f_n\|_p\to\|f\|_p$.

This is can be easily proved by the triangle inequality: $$ |\|f_n\|_p-\|f\|_p|\leq \|f_n-f\|_p. $$

Again, in $$ \lim_{n\to \infty} \langle f_n, g \rangle = \langle \lim_{n\to \infty} f_n, g \rangle = \langle f, g \rangle,\tag{2} $$ the notation $\lim_{n\to \infty} f_n=f$ should be understood as convergence in the Hilbert space, namely, convergence in norm induced by the inner product. It is a simple exercise by Cauchy-Schwarz: $$ |\langle f_n,g\rangle-\langle f,g\rangle|\leq \|f_n-f\|\cdot \|g\|. $$