# Uniform convergence of sequence of convex functions

Let $\{f_n\}$ converge point-wise to $f$, where each $f_n:[a,b]\rightarrow \mathbb{R}$, and each $f_n$ is a continuous convex function. Furthermore, assume that $f$ is continuous. Prove that the convergence is uniform.

I was trying to do something like: Consider $a=a_0<a_i<...<a_k=b$, where $a_{i+1}-a_i<\delta$, where the $\delta$ is such that $|x-y|<\delta$ imply that $|f(x)-f(y)|<\epsilon$. Then letting $N_i$ be such that $f_n(a_i)$ is $\epsilon$-close to $f(x_i)$, let $N$ be the max of $N_1,...,N_k$, and hence$$|f(x)-f_n(x)|\leq |f(x)-f(x_i)|+|f(x_i)-f_n(x_i)|+|f_n(x_i)-f_n(x)|$$I can make the first summand small since $f$ is continuos, and the second summand small by letting $n\geq N$. I am having trouble using the fact that they are convex.

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Suppose that the convergence is not uniform. Then there exists $\epsilon > 0$ and for every $n\in\mathbb{N}$, there exists $x_n\in [a,b]$ such that $|f_n(x_n) - f(x_n)|> \epsilon$. By compactness of $[a,b]$, the sequence $\{x_n\}$ has an accumulation point in $[a, b]$ - call this point $x$ (that is, there is a subsequence $\{x_{n_k}\}$ which converges to $x$). By pointwise convergence of $\{f_n\}$ to $f$, we know that for some $N\in\mathbb{N}$ and all $k\geq N$, $|f_{n_k}(x) - f(x)| < \epsilon/4$.

Since $f$ is continuous, it is uniformly continuous. Let $\delta > 0$ be such that for all $y_1, y_2\in [x - \delta, x + \delta]$, $|f(y_1) - f(y_2)| < \epsilon/2$. Let $\tilde{x}\in[a,b]$ with $\tilde{x} < x$ (assuming $x\neq a$; otherwise, take $\tilde{x} > x$), and $|\tilde{x}-x|< \delta$. Assume also that there are infinitely many terms of the subsequence $\{x_{n_k}\}$ which lie in $[\tilde{x}, x]$. But then we have the following situation: for all $k$ sufficiently large, $|f_{n_k}(x) - f(x)| < \epsilon/4$, $|f_{n_k}(\tilde{x}) - f(\tilde{x})| < \epsilon/4$ and there exists $y\in [\tilde{x}, x]$ with $|f_{n_k}(y) - f(y)| > \epsilon$. This then violates convexity of $f_{n_k}$ (to see this, consider lines connecting $\tilde{x}$ to $x_{n_k}$, one connecting $x_{n_k}$ to $x$, and one connecting $x$ to $\tilde{x}$ - one of these will lie below the graph of $f_{n_k}$).

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 Do you want to respond to Bernard's query below? – user31373 Sep 13 '12 at 20:21

I was trying to go through William's argument, but I got a bit confused towards the end. My guess is William was trying to show that by convexity that $|f_{n_k}(x_{n_k})-f(x_{n_k})|< \epsilon$ thus contradicting the construction of the sequence $x_n$. So, my approach to that was to try to prove that $f_{n_k}(x_{n_k})-f(x_{n_k})< \epsilon$ and also $f(x_{n_k})-f_{n_k}(x_{n_k})< \epsilon$. The first follows easily by convexity of $f_{n_k}$ and uniform continuity of $f$.

It is the second inequality I am having trouble with.

If we try to use convexity for $f_{n_k}$ we get nowhere since that leads to the first inequality. Thus we are almost forced to use convexity of $f$. But then, we get at some point on the RHS the terms $|f_{n_k}(x_{n_k})-f_{n_k}(x)|$ and $|f_{n_k}(x_{n_k})-f_{n_k}(y)|$.

But how can we bound these termes by some multiple of $\epsilon$? There is no guarantee that $x_{n_k}$ are close enough to x or y for the continuity of $f_{n_k}$ to work; this would essentially imply that the sequence is equicontinuous, so then we would be done by Arzela-Ascoli, right?

Is there something I am missing here?

Or is the approach to the second inequality completely different than the above?

Thanks in advance for any help...

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 No, that's not what I was trying to do. If $f_{n_k}(y) > f(y)$, then we are done. But it may happen that $f_{n_k} < f(y)$, then one needs to partition the interval $[\tilde{x}, x]$ into sufficiently small subintervals such that at each endpoint, $f_{n_k}$ is within $\epsilon/4$ of $f$ (for $k$ sufficiently large, of course) and then again use the fact that $f_{n_k}(y)$ is farther than $\epsilon$ from $f(y)$ for some $y$ in $(\tilde{x},x)$. – William Sep 14 '12 at 2:26 Sorry, I still fail to see how you do that. If you divide $[y,x]$ so that $f_{n_k}$ is within $\epsilon /4$ of $f$ at the endpoints, then you can't guarantee that $f_{n_k}(x_{n_k})$ will be within $\epsilon /4$ of the value of $f_{n_k}$ at the endpoints of the subinterval containing $x_{n_k}$. If you fix $n_k$, and pick the lengths of the interval by uniform continuity for $f_{n_k}$ then you do cannot guarantee that at the endpoints $f_{n_k}$ is within $\epsilon /4$ of f. Unless you use a trick or something; in which case, could you explain more? Thanks. – Bernard Sep 15 '12 at 5:21 I don't want to guarantee that $f_{n_k}(x_{n_k})$ will be within $\epsilon/4$ of the value of $f_{n_k}$ at the endpoints. I want just the opposite: $f_{n_k}(x_{n_k})$ is distance at least $\epsilon$ away from $f$, hence at least $3\epsilon/4$ away from the value of $f_{n_k}$ at the endpoints. This would force the graph of $f_{n_k}$ at at least one of those endpoints to lie above either the line connecting $(y, 0)$ to $(x_{n_k}, f_{n_k}(x_{n_k}))$, or the line connecting $(x, 0)$ to $(x_{n_k}, f_{n_k}(x_{n_k}))$ (assuming $f_{n_k}(x_{n_k}) < f(x_{n_k})$, violating convexity of $f_{n_k}$. – William Sep 15 '12 at 8:09 Occasionally, comments get too long or complex, and they need to be posted as answers; I have done this several times. However, to keep flags from being raised, I try to preface my comment-answers with something like: "this was intended to be a comment to someone's question/answer, but it was too long." – robjohn♦ May 10 at 15:29