Does uniform convergence in $L^1$ imply uniform convergence in $L^2$?

Question

I had to keep the title reasonably long, so below are the full details. This is a question that came to my mind while working on a distantly related problem; since I do not know whether the answer is negative or affirmative, I have no idea whether I should look for counterexamples or, on the contrary, I should try to come up with a proof.

Let $X$ be a space endowed with a finite measure $m$ (also, there exist non-empty subsets of measure $0$, to prevent some trivial counterexamples). Let $T$ be some compact Hausdorff topological space. Let $f_k : T \times X \to \mathbb R$ be a family of functions indexed by $k \in \mathbb N$ such that:

• $f_k (t, \cdot) \in L^2 (X) \subseteq L^1(X)$ for all $k \ge 0$;
• the map $T \ni t \mapsto f_k (t, \cdot) \in L^1 (X)$ is continuous with respect to the $L^2$ norm;
• $f_k (t, \cdot) \to 0$ in $L^1(X)$, uniformly with respect to $t \in T$;
• $f_k (t, \cdot) \to 0$ in $L^2(X)$ for every $t \in T$.

Is the convergence in $L^2$ also uniform with respect to $t \in T$, like the convergence in $L^1$?

Answer 1

The answer is negative.

Consider the space $X = \left( [0, 1], \mathcal{B}\left( [0, 1] \right), \lambda \vert_{[0, 1]} \right)$, where $\mathcal{B}\left( [0, 1 ] \right)$ denotes the Borel $\sigma$-algebra and $\lambda \vert_{[0, 1]}$ denotes the restriction of the Lebesgue measure. Consider also the space $T = [0, 1]$ endowed with its usual topology. Given an integer $k \geq 1$, define the map $f_{k} \colon T \times X \rightarrow \mathbb{R}$ by $$f_{k} \colon (t, x) \mapsto \begin{cases} \max\left\lbrace k \left( 1 -\lvert k t -1 \rvert \right), 0 \right\rbrace & \text{if } x \in \left[ \frac{1}{k^{2}}, \frac{2}{k^{2}} \right]\\ 0 & \text{otherwise} \end{cases} \, \text{.}$$

The desired conditions are satisfied:

• For every integer $k \geq 1$ and every $t \in T$, we have $f_{k}(t, .) \in L^{2}(X)$ since $f_{k}(t, .)$ is bounded.
• For every integer $k \geq 1$ and every $t_{0} \in T$, the map $f_{k}(t, .)$ tends to $f_{k}\left( t_{0}, . \right)$ in $L^{\infty}(X)$ as $t \rightarrow t_{0}$. Therefore, for every $k \geq 1$, the map $t \in T \mapsto f_{k}(t, .) \in L^{2}(X)$ is continuous.
• For every integer $k \geq 1$ and every $t \in T$, we have $$\left\lVert f_{k}(t, .) \right\rVert_{L^{1}(X)} = \int_{\frac{1}{k^{2}}}^{\frac{2}{k^{2}}} \max\left\lbrace k \left( 1 -\lvert k t -1 \rvert \right), 0 \right\rbrace \, dx \leq \int_{\frac{1}{k^{2}}}^{\frac{2}{k^{2}}} k \, dx = \frac{1}{k} \, \text{.}$$ Therefore, we have $$\lim_{k \rightarrow +\infty} \sup_{t \in T} \left\lVert f_{k}(t, .) \right\rVert_{L^{1}(X)} = 0 \, \text{.}$$
• We have $f_{k}(0, .) = 0$ for all $k \geq 1$, and hence $f_{k}(0, .)$ tends to $0$ in $L^{2}(X)$ as $k \rightarrow +\infty$. For every $t \in ]0, 1]$, we have $f_{k}(t, .) = 0$ for all integers $k \geq \frac{2}{t}$, and hence $f_{k}(t, .)$ tends to $0$ in $L^{2}(X)$ as $k \rightarrow +\infty$.

For every integer $k \geq 1$, we have $$\left\lVert f_{k}\left( \frac{1}{k}, . \right) \right\rVert_{L^{2}(X)}^{2} = \int_{\frac{1}{k^{2}}}^{\frac{2}{k^{2}}} k^{2} \, dx = 1 \, \text{.}$$