Problem 1: Consider a bounded domain U with C² boundary. Let f denote a harmonic function in C²(U) that is continuous up to the boundary of U, which we denote by S. Let T be a subset of S, and suppose both f and the normal derivative of f vanish identically on T. Show that f is identically 0 on U.

Proof (for which I am indebted to Spyros): The case in 2 dimensions is obvious: consider the function grad f. It represents a complex analytic function on U. Continuity up to boundary guarantees an analytic extension to a larger bounded set V, which, by the Riemann mapping theorem has a holomorphic map to the unit disc. So, by the pull back, we can safely assume that we are dealing with a complex analytic function f on the unit disc that evaluates to zero on a line segment containing the origin, which implies that it has an accumulation of zeroes, and must equal zero throughout.

The case in higher dimensions uses the formulation of weak solutions. We extend U to a larger domain V by "protruding" a piece outside of T. Extend f to a function on V by setting Due to the fact that the normal derivative of f on T is 0, f_V is a W^1,2 function, and weakly satisfies the harmonicity condition on V. But it is know that a weakly harmonic function is also a strongly harmonic one, and we know that any harmonic function that is identically zero on an open subset must be identically zero. So f is zero on U.

Problem 2: Let {f_n} denote a sequence of continuous functions on the interval [0,1] converging point-wise to a function f, which has L¹ norm 0. Is f necessarily continuous? If not, does there exists points in [0,1] on which f is continuous?

Answer: f doesn't have to be continuous. Let f_n be defined as follows: They are a family of continuous functions, and point-wise converges to which is not continuous.

The answer to the second part is affirmative, however, and here I give a proof to that fact.

We start by considering the following sets We notice that A_N,ε are closed sets, and A_n,ε is contained in A_n+1,ε. Further more, for fixed ε, by point-wise convergence, each x in [0,1] belongs to some A_N,ε. So we have that the union of all A_N,ε is [0,1]. We can then apply the Baire Category Theorem to this countable collection and obtain that at least one of A_N,ε must be second Baire category. Since A_N,ε are closed, being in second Baire category is equivalent to saying that the set of interior points is non-empty.

Let V_N,ε be the interior points of A_N,ε. It is by definition open, and for each fixed ε there exists at least one N such that V_N,ε is non-empty. I claim that for each fixed ε, V_ε, the union over N of all V_N,ε, is dense and open in [0,1]. The open property is obvious by countable union. It suffices to show density.

Suppose V_ε is not dense in [0,1], then its complement in [0,1] is closed and contains an interval, which makes it second category. Let B_N denote the intersection of V_ε^C with A_N,ε. Baire Category implies one of B_N is second category, and contains an interior point. By definition, the interior point must be contained in V_ε, contradicting the assumption.

Consider the family V_1/k where k is a natural number. It is a countable family of open and dense subsets of [0,1], and by the Baire Category Theorem again we have that its intersection V is dense in [0,1].

We finish by showing that the function f must be continuous on V.

Let x be a point in V. For each ε₀ > 0, we set ε = ε₀/3. Choose k such that 1/k < ε. Then x also belongs to V_1/k. Which implies that there exists some N_ε(x) and an open neighborhood B_ε(x) such that Choose some m > N_ε(x). By continuity of f_m, there exists a neighborhood U(x) s.t. Then, for y in the open neighborhood U(x) intersect B_ε(x), we have So f is continuous at x.