Does convergence in H1 imply pointwise convergence?

I'm trying to figure out if convergence in $H^1(a,b)$ implies pointwise convergence (by the way: what is the usual name of this space?). It is defined to be Hilbert space of absolutely continuous functions on some (possibly infinite) interval such that both function and its (weak) derivative are square integrable. Scalar product in this space is $(f,g)_{H^1}=(f,g)_{L^2}+(f',g')_{L^2}$. I found out thar if we know that $f_n$ converges to $f$ in $H^1$ and for some $c \in (a,b)$ limit of sequence of numbers $f_n(c)$ exists then $f_n$ converges pointwise (I post the proof below).

Edit: I add the proof of claim given above. Suppose $f_n(c) \to g$ and $f_n \to f$ in $H^1$. We have

$f_n(x)=f_n(c)+\int _c ^xf'_n(t)dt \to g + \int _c ^xf'(t) dt$

$f_n$ has sequence converging a.e. to $f$. By uniqueness of limits and continuity of $f$ it is easy to see that $g=f(c)$ so $f_n(x) \to f(x)$. This argument also shows that even without assumption of convergence in one point $f_n$ must have subsequence convergent pointwise everywhere.

By the way it would be odd to me if sequence converging in $L^2$ existed that doesn't converge pointwise in any point. Can anyone give example?

One can complete your argument using the "subsequence principle". I.e. in a topological space, we have $x_n \to x$ iff every subsequence $(x_{n_k})_k$ has a further subsequence $(x_{n_{k_\ell}})_\ell$ with $x_{n_{k_\ell}} \to x$. Note though, that this only holds if the limit $x$ is fixed beforehand.

Fix some $x_0 \in (a,b)$. There are now two cases:

1. $(|f_n (x_0)|)_n$ is an unbounded sequence. By choosing a subsequence, we can assume $|f_n (x_0)| \to \infty$ as $n \to \infty$. Using your argument, the "reverse triangle inequality", and the Cauchy-Schwarz inequality, we get \begin{eqnarray*} |f_n (x)| &=& |f_n (x_0) + \int_{x_0}^x f_n ' (t) \ dt| \geq |f_n (x_0)| - \int_a^b |f_n '(t)| \, dt \\ &\geq &|f_n(x_0)| - \sqrt{b-a} \cdot \sqrt{\int_a^b |f_n'(t)|^2 \, dt}\\ & \geq & |f_n (x_0)| - C \end{eqnarray*} and hence (choosing $n$ so large that $|f_n(x_0)| \geq C$): $$\int |f_n (x)|^2 \, dx \geq \int (|f_n(x_0)| - C)^2 \, dx = (b-a) \cdot (|f_n(x_0)| - C)^2 \to \infty,$$ a contradiction.

2. The sequence $(f_n(x_0))_n$ is bounded. Assume that $f_n (x) \to f(x)$ does not hold for all $x$. Let $y$ be a point where this convergence does not hold. Then there is $\varepsilon > 0$ and a subsequence $(g_k)_k = (f_{n_k})_k$ of $(f_n)_n$ with $|g_k (y) - f(y)| \geq \varepsilon$ for all $k$.

Since $(g_k (x_0))_k$ is bounded, we can choose a subsequence $(g_{k_\ell})_\ell$, so that $g_{k_\ell} (x_0)$ converges to some $g \in \Bbb{R}$. As you noted, this implies $g_{k_\ell}(x) \to f(x)$ for all $x$. In particular, $g_{k_\ell} (y) \to f(y)$, which contradicts $|g_k (y) - f(y)| \geq \varepsilon$ for all $k$. This contradiction shows that $f_n (x) \to f(x)$ for all $x$ must hold.

Regaring the counterexample for $L^2$ convergence, consider the sequence

$$\chi_{[0,1]}, \chi_{[0,1/2]}, \chi_{[1/2, 1]}, \chi_{[0,1/3]}, \chi_{[1/3, 2/3]}, \chi_{[2/3, 1]}, \dots.$$ It is not hard to see that $f_n(x)$ accumulates at $0$ and at $1$ at every $x$, so that $f_n(x)$ converges for no $x$, but still $\Vert f_n \Vert_2 \to 0$.

EDIT: Once one has pointwise convergence, it is not hard to see that the identity $f_n (x) = f_n (x_0) + \int_{x_0}^x f_n '(t) \, dt$ even implies uniform convergence.

• Thank you for great answer. Analysing it led me to great simplification of proof: I already mentioned that every sequence converging in H^1 has subsequence converging pointwise to its H1 limit everywhere. Therefore every subsequence of such sequence also has subsequence with this property. Pointwise convergence follows. It is interesting to note that both this and uniform convergence also hold for infinite intervals so H1(R) has topology quite stronger than one might think at first sight. – Blazej Jul 9 '15 at 15:33
• @Blazej: oh yes, this simplifies the argument quite a bit. Note though that the convergence is pointwise but not uniform on unbounded intervals (I think). At least the proof breaks down, because you can not estimate $\int |f'|dx$ by $\sqrt{\int |f'|^2}$. – PhoemueX Jul 9 '15 at 17:50
• Yes you are right, it breaks down. I've made a simple mistake in my calculations, thanks for pointing it out. – Blazej Jul 9 '15 at 22:28

The space $H^1$ is a short notation for $W^{1,2}$ which is one of Sobolev spaces. (Not very good notation in general since it can be mixed up with the Hardy space which is something completely different, so I prefer to use $W^{1,2}$.)

The result you are asking for follows easily from the Morrey's inequality. It tells us, in particular, that for $n=k=1$ and $p=2$ the space $W^{1,2}(a,b)$ can be continuously embedded into the Hölder space $C^{0,1/2}(a,b)$, i.e. the original convergence implies the convergence in the $1/2$-Hölder norm, which is (by the norm definition in $C^{0,1/2}$) stronger than the uniform convergence.