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I give a proof that gives tighter sup-norm bounds when you have control of the $L^2$ norm (or $L^q$ norm more generally). The following proof should also be generalizable to general Holder classes (e.g. functions with Lipschitz derivatives). I give the rigorous proof for $d=1$ as the generalization to dimension $d$ follows easily.

Dimension 1

Suppose that $f_n$ and $f_0$ are L-Lipschitz on the interval $I:=[A,B]$. Let $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$.

The key idea of the proof is the following proposition. Note that I implicitly assume $\frac{\varepsilon_n}{4L} \leq (B-A)$. With some minor modifications, the more general case can also be handled.

Proposition (1). Suppose $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$. Then, there exists there exists some $c \in [A, B - \delta_n]$ such that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \geq \varepsilon_n/2$ where $\delta_n := \frac{\varepsilon_n}{4L}$.

Proof. It suffices to prove the following contrapositive: If for all $c \in [A, B - \delta_n]$ we have $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$ then $ \|f_n-f_0\|_{\infty, I} \leq \varepsilon$.

To this end, suppose that for all $c \in [A, B - \delta_n]$ that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$. Take some $c \in [A, B - \delta_n]$ and let $t_n \in [c,c+\delta_n]$ be such that $ |f_n(t_n) - f_0(t_n)| \leq \varepsilon_n/2$, which we can do by assumption. Then, $$\text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_n(t_n)| + |f_n(t_n) - f_0(t_n)| + \text{sup}_{x \in [c,c+\delta_n]} |f_0(t_n) - f_0(x)| \leq 2L\delta_n + \varepsilon_n/2,$$ where the RHS bound holds independent of $c$ and $t_n$. Thus, we actually have $$\text{sup}_{x \in I} |f_n(x) - f_0(x)| \leq 2L\delta_n + \varepsilon_n /2\leq \varepsilon_n$$ by our choice of $\delta_n$. This proves the claim. (One can prove similar propositions for Holder classes more generally (e.g. functions with Lipschitz derivative.) End Proof.

Let $B_n = [c,c+\delta_n]$ be the interval guaranteed by Proposition 1. with $\inf_{x \in B_n}|f_n(x) - f_0(x)| \geq \|f_n-f_0\|_{\infty} = \varepsilon_n$ (where we will ignore constants for simplicity). It follows that $$ \varepsilon_n^{2} \lessapprox {\delta_n} \varepsilon_n \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{1/2},$$ and $$ \varepsilon_n^{3/2} \lessapprox \sqrt{\delta_n} \varepsilon_n \lessapprox \sqrt{{\int_{B_n} |f_n-f_0|^2(x) dx}} \leq \|f_n - f_0\|_{L^2}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{2/3}.$$ More generally, we can show for $q > 0$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{q+1}}.$$

Higher dimensions

Proposition (1) can be generalized to dimension $d$ with virtually no changes but an addition of a constant $O(\sqrt{d})$ in some places. In particular, one can show that $\|f_n-f_0\|_{\infty} = C_1\varepsilon_n$ implies there is a rectangle $B_n$ with sides at most $C_2\varepsilon_n$ such that $\inf_{x\in B_n}|f_n(x)-f_0(x)| \geq C_3 \varepsilon_n$ for constants $C_i$>0. Thus, $$ \varepsilon_n \varepsilon_n^{ d} \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ and $$ \varepsilon_n \varepsilon_n^{d/2 } \lessapprox \sqrt{\int_{B_n} |f_n-f_0|(x)^2 dx} \leq \|f_n - f_0\|_{L^2}$$ which implies $$ \|f_n-f_0\|_{\infty}\lessapprox \|f_n - f_0\|_{L^1}^{\frac{1}{1+d}}$$ and $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{2}{d+2}}.$$ More generally, we can show for $q > 0$ $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}.$$ A simple but interesting consequence of the above is as follows. Since $$ \|f_n - f_0\|_{L^p} \lessapprox \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}},$$ we have $$ \|f_n - f_0\|_{L^q}^{\frac{d+p}{p}} \lessapprox \|f_n - f_0\|_{L^p} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}$$ for all $p,q>0$.

Higher-order smoothness

If one has that $f_n$ and $f_0$ are univariate Lipschitz with $L$-Lipschitz derivatives, one can show by a similar argument (proving a tighter version of proposition (1) with first order taylor expansions and with $\delta_n = C\varepsilon_n^{1/2}$) that $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{\frac{2}{3}}.$$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{4}{5}}.$$ More generally, we can show for dimension $d$ and functions with $(\alpha-1)$-order derivative being $L$-Lipschitz that $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q\alpha}{d+q\alpha}}.$$

I give a proof that gives tighter sup-norm bounds when you have control of the $L^2$ norm (or $L^q$ norm more generally). The following proof should also be generalizable to general Holder classes (e.g. functions with Lipschitz derivatives). I give the rigorous proof for $d=1$ as the generalization to dimension $d$ follows easily.

Dimension 1

Suppose that $f_n$ and $f_0$ are L-Lipschitz on the interval $I:=[A,B]$. Let $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$.

The key idea of the proof is the following proposition.

Proposition (1). Suppose $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$. Then, there exists there exists some $c \in [A, B - \delta_n]$ such that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \geq \varepsilon_n/2$ where $\delta_n := \frac{\varepsilon_n}{4L}$.

Proof. Note that I implicitly assume $\frac{\varepsilon_n}{4L} \leq (B-A)$. With some minor modifications, the proposition remains true for the general case with different constants.

It suffices to prove the following contrapositive: If for all $c \in [A, B - \delta_n]$ we have $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$ then $ \|f_n-f_0\|_{\infty, I} \leq \varepsilon$.

To this end, suppose that for all $c \in [A, B - \delta_n]$ that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$. Take some $c \in [A, B - \delta_n]$ and let $t_n \in [c,c+\delta_n]$ be such that $ |f_n(t_n) - f_0(t_n)| \leq \varepsilon_n/2$, which we can do by assumption. Then, $$\text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_n(t_n)| + |f_n(t_n) - f_0(t_n)| + \text{sup}_{x \in [c,c+\delta_n]} |f_0(t_n) - f_0(x)| \leq 2L\delta_n + \varepsilon_n/2,$$ where the RHS bound holds independent of $c$ and $t_n$. Thus, we actually have $$\text{sup}_{x \in I} |f_n(x) - f_0(x)| \leq 2L\delta_n + \varepsilon_n /2\leq \varepsilon_n$$ by our choice of $\delta_n$. This proves the claim. (One can prove similar propositions for Holder classes more generally (e.g. functions with Lipschitz derivative.) End Proof.

Let $B_n = [c,c+\delta_n]$ be the interval guaranteed by Proposition 1. with $\inf_{x \in B_n}|f_n(x) - f_0(x)| \geq \|f_n-f_0\|_{\infty} = \varepsilon_n$ (where we will ignore constants for simplicity). It follows that $$ \varepsilon_n^{2} \lessapprox {\delta_n} \varepsilon_n \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{1/2},$$ and $$ \varepsilon_n^{3/2} \lessapprox \sqrt{\delta_n} \varepsilon_n \lessapprox \sqrt{{\int_{B_n} |f_n-f_0|^2(x) dx}} \leq \|f_n - f_0\|_{L^2}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{2/3}.$$ More generally, we can show for $q > 0$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{q+1}}.$$

Higher dimensions

Proposition (1) can be generalized to dimension $d$ with virtually no changes but an addition of a constant $O(\sqrt{d})$ in some places. In particular, one can show that $\|f_n-f_0\|_{\infty} = C_1\varepsilon_n$ implies there is a rectangle $B_n$ with sides at most $C_2\varepsilon_n$ such that $\inf_{x\in B_n}|f_n(x)-f_0(x)| \geq C_3 \varepsilon_n$ for constants $C_i$>0. Thus, $$ \varepsilon_n \varepsilon_n^{ d} \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ and $$ \varepsilon_n \varepsilon_n^{d/2 } \lessapprox \sqrt{\int_{B_n} |f_n-f_0|(x)^2 dx} \leq \|f_n - f_0\|_{L^2}$$ which implies $$ \|f_n-f_0\|_{\infty}\lessapprox \|f_n - f_0\|_{L^1}^{\frac{1}{1+d}}$$ and $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{2}{d+2}}.$$ More generally, we can show for $q > 0$ $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}.$$ A simple but interesting consequence of the above is as follows. Since $$ \|f_n - f_0\|_{L^p} \lessapprox \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}},$$ we have $$ \|f_n - f_0\|_{L^q}^{\frac{d+p}{p}} \lessapprox \|f_n - f_0\|_{L^p} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}$$ for all $p,q>0$.

Higher-order smoothness

If one has that $f_n$ and $f_0$ are univariate Lipschitz with $L$-Lipschitz derivatives, one can show by a similar argument (proving a tighter version of proposition (1) with first order taylor expansions and with $\delta_n = C\varepsilon_n^{1/2}$) that $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{\frac{2}{3}}.$$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{4}{5}}.$$ More generally, we can show for dimension $d$ and functions with $(\alpha-1)$-order derivative being $L$-Lipschitz that $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q\alpha}{d+q\alpha}}.$$

I give a proof that gives tighter sup-norm bounds when you have control of the $L^2$ norm (or $L^q$ norm more generally). The following proof should also be generalizable to general Holder classes (e.g. functions with Lipschitz derivatives). I give the rigorous proof for $d=1$ as the generalization to dimension $d$ follows easily.

Dimension 1

Suppose that $f_n$ and $f_0$ are L-Lipschitz on the interval $I:=[A,B]$. Let $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$.

The key idea of the proof is the following proposition.

Proposition (1). Suppose $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$. Then, there exists there exists some $c \in [A, B - \delta_n]$ such that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \geq \varepsilon_n/2$ where $\delta_n := \frac{\varepsilon_n}{4L}$.

Proof. It suffices to prove the following contrapositive: If for all $c \in [A, B - \delta_n]$ we have $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$ then $ \|f_n-f_0\|_{\infty, I} \leq \varepsilon$.

To this end, suppose that for all $c \in [A, B - \delta_n]$ that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$. Take some $c \in [A, B - \delta_n]$ and let $t_n \in [c,c+\delta_n]$ be such that $ |f_n(t_n) - f_0(t_n)| \leq \varepsilon_n/2$, which we can do by assumption. Then, $$\text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_n(t_n)| + |f_n(t_n) - f_0(t_n)| + \text{sup}_{x \in [c,c+\delta_n]} |f_0(t_n) - f_0(x)| \leq 2L\delta_n + \varepsilon_n/2,$$ where the RHS bound holds independent of $c$ and $t_n$. Thus, we actually have $$\text{sup}_{x \in I} |f_n(x) - f_0(x)| \leq 2L\delta_n + \varepsilon_n /2\leq \varepsilon_n$$ by our choice of $\delta_n$. This proves the claim. (One can prove similar propositions for Holder classes more generally (e.g. functions with Lipschitz derivative.) End Proof.

Let $B_n = [c,c+\delta_n]$ be the interval guaranteed by Proposition 1. with $\inf_{x \in B_n}|f_n(x) - f_0(x)| \geq \|f_n-f_0\|_{\infty} = \varepsilon_n$ (where we will ignore constants for simplicity). It follows that $$ \varepsilon_n^{2} \lessapprox {\delta_n} \varepsilon_n \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{1/2},$$ and $$ \varepsilon_n^{3/2} \lessapprox \sqrt{\delta_n} \varepsilon_n \lessapprox \sqrt{{\int_{B_n} |f_n-f_0|^2(x) dx}} \leq \|f_n - f_0\|_{L^2}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{2/3}.$$ More generally, we can show for $q > 0$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{q+1}}.$$

Higher dimensions

Proposition (1) can be generalized to dimension $d$ with virtually no changes but an addition of a constant $O(\sqrt{d})$ in some places. In particular, one can show that $\|f_n-f_0\|_{\infty} = C_1\varepsilon_n$ implies there is a rectangle $B_n$ with sides at most $C_2\varepsilon_n$ such that $\inf_{x\in B_n}|f_n(x)-f_0(x)| \geq C_3 \varepsilon_n$ for constants $C_i$>0. Thus, $$ \varepsilon_n \varepsilon_n^{ d} \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ and $$ \varepsilon_n \varepsilon_n^{d/2 } \lessapprox \sqrt{\int_{B_n} |f_n-f_0|(x)^2 dx} \leq \|f_n - f_0\|_{L^2}$$ which implies $$ \|f_n-f_0\|_{\infty}\lessapprox \|f_n - f_0\|_{L^1}^{\frac{1}{1+d}}$$ and $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{2}{d+2}}.$$ More generally, we can show for $q > 0$ $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}.$$ A simple but interesting consequence of the above is as follows. Since $$ \|f_n - f_0\|_{L^p} \lessapprox \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}},$$ we have $$ \|f_n - f_0\|_{L^q}^{\frac{d+p}{p}} \lessapprox \|f_n - f_0\|_{L^p} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}$$ for all $p,q>0$.

Higher-order smoothness

If one has that $f_n$ and $f_0$ are univariate Lipschitz with $L$-Lipschitz derivatives, one can show by a similar argument (proving a tighter version of proposition (1) with first order taylor expansions and with $\delta_n = C\varepsilon_n^{1/2}$) that $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{\frac{2}{3}}.$$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{4}{5}}.$$ More generally, we can show for dimension $d$ and functions with $(\alpha-1)$-order derivative being $L$-Lipschitz that $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q\alpha}{d+q\alpha}}.$$

I give a proof that gives tighter sup-norm bounds when you have control of the $L^2$ norm (or $L^q$ norm more generally). The following proof should also be generalizable to general Holder classes (e.g. functions with Lipschitz derivatives). I give the rigorous proof for $d=1$ as the generalization to dimension $d$ follows easily.

Dimension 1

Suppose that $f_n$ and $f_0$ are L-Lipschitz on the interval $I:=[A,B]$. Let $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$.

The key idea of the proof is the following proposition. Note that I implicitly assume $\frac{\varepsilon_n}{4L} \leq (B-A)$. With some minor modifications, the more general case can also be handled.

Proposition (1). Suppose $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$. Then, there exists there exists some $c \in [A, B - \delta_n]$ such that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \geq \varepsilon_n/2$ where $\delta_n := \frac{\varepsilon_n}{4L}$.

Proof. It suffices to prove the following contrapositive: If for all $c \in [A, B - \delta_n]$ we have $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$ then $ \|f_n-f_0\|_{\infty, I} \leq \varepsilon$.

To this end, suppose that for all $c \in [A, B - \delta_n]$ that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$. Take some $c \in [A, B - \delta_n]$ and let $t_n \in [c,c+\delta_n]$ be such that $ |f_n(t_n) - f_0(t_n)| \leq \varepsilon_n/2$, which we can do by assumption. Then, $$\text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_n(t_n)| + |f_n(t_n) - f_0(t_n)| + \text{sup}_{x \in [c,c+\delta_n]} |f_0(t_n) - f_0(x)| \leq 2L\delta_n + \varepsilon_n/2,$$ where the RHS bound holds independent of $c$ and $t_n$. Thus, we actually have $$\text{sup}_{x \in I} |f_n(x) - f_0(x)| \leq 2L\delta_n + \varepsilon_n /2\leq \varepsilon_n$$ by our choice of $\delta_n$. This proves the claim. (One can prove similar propositions for Holder classes more generally (e.g. functions with Lipschitz derivative.) End Proof.

Let $B_n = [c,c+\delta_n]$ be the interval guaranteed by Proposition 1. with $\inf_{x \in B_n}|f_n(x) - f_0(x)| \geq \|f_n-f_0\|_{\infty} = \varepsilon_n$ (where we will ignore constants for simplicity). It follows that $$ \varepsilon_n^{2} \lessapprox {\delta_n} \varepsilon_n \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{1/2},$$ and $$ \varepsilon_n^{3/2} \lessapprox \sqrt{\delta_n} \varepsilon_n \lessapprox \sqrt{{\int_{B_n} |f_n-f_0|^2(x) dx}} \leq \|f_n - f_0\|_{L^2}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{2/3}.$$ More generally, we can show for $q > 0$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{q+1}}.$$

Higher dimensions

Proposition (1) can be generalized to dimension $d$ with virtually no changes but an addition of a constant $O(\sqrt{d})$ in some places. In particular, one can show that $\|f_n-f_0\|_{\infty} = C_1\varepsilon_n$ implies there is a rectangle $B_n$ with sides at most $C_2\varepsilon_n$ such that $\inf_{x\in B_n}|f_n(x)-f_0(x)| \geq C_3 \varepsilon_n$ for constants $C_i$>0. Thus, $$ \varepsilon_n \varepsilon_n^{ d} \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ and $$ \varepsilon_n \varepsilon_n^{d/2 } \lessapprox \sqrt{\int_{B_n} |f_n-f_0|(x)^2 dx} \leq \|f_n - f_0\|_{L^2}$$ which implies $$ \|f_n-f_0\|_{\infty}\lessapprox \|f_n - f_0\|_{L^1}^{\frac{1}{1+d}}$$ and $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{2}{d+2}}.$$ More generally, we can show for $q > 0$ $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}.$$ A simple but interesting consequence of the above is as follows. Since $$ \|f_n - f_0\|_{L^p} \lessapprox \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}},$$ we have $$ \|f_n - f_0\|_{L^q}^{\frac{d+p}{p}} \lessapprox \|f_n - f_0\|_{L^p} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}$$ for all $p,q>0$.

Higher-order smoothness

If one has that $f_n$ and $f_0$ are univariate Lipschitz with $L$-Lipschitz derivatives, one can show by a similar argument (proving a tighter version of proposition (1) with first order taylor expansions and with $\delta_n = C\varepsilon_n^{1/2}$) that $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{\frac{2}{3}}.$$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{4}{5}}.$$ More generally, we can show for dimension $d$ and functions with $(\alpha-1)$-order derivative being $L$-Lipschitz that $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q\alpha}{d+q\alpha}}.$$

I give a proof that gives tighter sup-norm bounds when you have control of the $L^2$ norm (or $L^q$ norm more generally). The following proof should also be generalizable to general Holder classes (e.g. functions with Lipschitz derivatives). I give the rigorous proof for $d=1$ as the generalization to dimension $d$ follows easily.

Dimension 1

Suppose that $f_n$ and $f_0$ are L-Lipschitz on the interval $I:=[A,B]$. Let $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$.

The key idea of the proof is the following proposition.

Proposition (1). Suppose $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$. Then, there exists there exists some $c \in [A, B - \delta_n]$ such that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \geq \varepsilon_n/2$ where $\delta_n := \frac{\varepsilon_n}{4L}$.

Proof. It suffices to prove the following contrapositive: If for all $c \in [A, B - \delta_n]$ we have $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$ then $ \|f_n-f_0\|_{\infty, I} \leq \varepsilon$.

To this end, suppose that for all $c \in [A, B - \delta_n]$ that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$. Take some $c \in [A, B - \delta_n]$ and let $t_n \in [c,c+\delta_n]$ be such that $ |f_n(t_n) - f_0(t_n)| \leq \varepsilon_n/2$, which we can do by assumption. Then, $$\text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_n(t_n)| + |f_n(t_n) - f_0(t_n)| + \text{sup}_{x \in [c,c+\delta_n]} |f_0(t_n) - f_0(x)| \leq 2L\delta_n + \varepsilon_n/2,$$ where the RHS bound holds independent of $c$ and $t_n$. Thus, we actually have $$\text{sup}_{x \in I} |f_n(x) - f_0(x)| \leq 2L\delta_n + \varepsilon_n /2\leq \varepsilon_n$$ by our choice of $\delta_n$. This proves the claim. (One can prove similar propositions for Holder classes more generally (e.g. functions with Lipschitz derivative.) End Proof.

Let $B_n = [c,c+\delta_n]$ be the interval guaranteed by Proposition 1. with $\inf_{x \in B_n}|f_n(x) - f_0(x)| \geq \|f_n-f_0\|_{\infty} = \varepsilon_n$ (where we will ignore constants for simplicity). It follows that $$ \varepsilon_n^{2} \lessapprox {\delta_n} \varepsilon_n \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{1/2},$$ and $$ \varepsilon_n^{3/2} \lessapprox \sqrt{\delta_n} \varepsilon_n \lessapprox \sqrt{{\int_{B_n} |f_n-f_0|^2(x) dx}} \leq \|f_n - f_0\|_{L^2}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{2/3}.$$ More generally, we can show for $q > 0$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{q+1}}.$$

Higher dimensions

Proposition (1) can be generalized to dimension $d$ with virtually no changes but an addition of a constant $O(\sqrt{d})$ in some places. In particular, one can show that $\|f_n-f_0\|_{\infty} = C_1\varepsilon_n$ implies there is a rectangle $B_n$ with sides at most $C_2\varepsilon_n$ such that $\inf_{x\in B_n}|f_n(x)-f_0(x)| \geq C_3 \varepsilon_n$ for constants $C_i$>0. Thus, $$ \varepsilon_n \varepsilon_n^{ d} \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ and $$ \varepsilon_n \varepsilon_n^{d/2 } \lessapprox \sqrt{\int_{B_n} |f_n-f_0|(x)^2 dx} \leq \|f_n - f_0\|_{L^2}$$ which implies $$ \|f_n-f_0\|_{\infty}\lessapprox \|f_n - f_0\|_{L^1}^{\frac{1}{1+d}}$$ and $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{2}{d+2}}.$$ More generally, we can show for $q > 0$ $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}.$$

Higher order smoothness

If one has that $f_n$ and $f_0$ are univariate Lipschitz with $L$-Lipschitz derivatives, one can show by a similar argument (proving a tighter version of proposition (1) with first order taylor expansions and with $\delta_n = C\varepsilon_n^{1/2}$) that $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{\frac{2}{3}}.$$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{4}{5}}.$$ More generally, we can show for dimension $d$ and functions with $(\alpha-1)$-order derivative being $L$-Lipschitz that $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q\alpha}{d+q\alpha}}.$$

I give a proof that gives tighter sup-norm bounds when you have control of the $L^2$ norm (or $L^q$ norm more generally). The following proof should also be generalizable to general Holder classes (e.g. functions with Lipschitz derivatives). I give the rigorous proof for $d=1$ as the generalization to dimension $d$ follows easily.

Dimension 1

Suppose that $f_n$ and $f_0$ are L-Lipschitz on the interval $I:=[A,B]$. Let $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$.

The key idea of the proof is the following proposition.

Proposition (1). Suppose $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$. Then, there exists there exists some $c \in [A, B - \delta_n]$ such that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \geq \varepsilon_n/2$ where $\delta_n := \frac{\varepsilon_n}{4L}$.

Proof. It suffices to prove the following contrapositive: If for all $c \in [A, B - \delta_n]$ we have $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$ then $ \|f_n-f_0\|_{\infty, I} \leq \varepsilon$.

To this end, suppose that for all $c \in [A, B - \delta_n]$ that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$. Take some $c \in [A, B - \delta_n]$ and let $t_n \in [c,c+\delta_n]$ be such that $ |f_n(t_n) - f_0(t_n)| \leq \varepsilon_n/2$, which we can do by assumption. Then, $$\text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_n(t_n)| + |f_n(t_n) - f_0(t_n)| + \text{sup}_{x \in [c,c+\delta_n]} |f_0(t_n) - f_0(x)| \leq 2L\delta_n + \varepsilon_n/2,$$ where the RHS bound holds independent of $c$ and $t_n$. Thus, we actually have $$\text{sup}_{x \in I} |f_n(x) - f_0(x)| \leq 2L\delta_n + \varepsilon_n /2\leq \varepsilon_n$$ by our choice of $\delta_n$. This proves the claim. (One can prove similar propositions for Holder classes more generally (e.g. functions with Lipschitz derivative.) End Proof.

Let $B_n = [c,c+\delta_n]$ be the interval guaranteed by Proposition 1. with $\inf_{x \in B_n}|f_n(x) - f_0(x)| \geq \|f_n-f_0\|_{\infty} = \varepsilon_n$ (where we will ignore constants for simplicity). It follows that $$ \varepsilon_n^{2} \lessapprox {\delta_n} \varepsilon_n \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{1/2},$$ and $$ \varepsilon_n^{3/2} \lessapprox \sqrt{\delta_n} \varepsilon_n \lessapprox \sqrt{{\int_{B_n} |f_n-f_0|^2(x) dx}} \leq \|f_n - f_0\|_{L^2}$$ which gives $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{2/3}.$$ More generally, we can show for $q > 0$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{q+1}}.$$

Higher dimensions

Proposition (1) can be generalized to dimension $d$ with virtually no changes but an addition of a constant $O(\sqrt{d})$ in some places. In particular, one can show that $\|f_n-f_0\|_{\infty} = C_1\varepsilon_n$ implies there is a rectangle $B_n$ with sides at most $C_2\varepsilon_n$ such that $\inf_{x\in B_n}|f_n(x)-f_0(x)| \geq C_3 \varepsilon_n$ for constants $C_i$>0. Thus, $$ \varepsilon_n \varepsilon_n^{ d} \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ and $$ \varepsilon_n \varepsilon_n^{d/2 } \lessapprox \sqrt{\int_{B_n} |f_n-f_0|(x)^2 dx} \leq \|f_n - f_0\|_{L^2}$$ which implies $$ \|f_n-f_0\|_{\infty}\lessapprox \|f_n - f_0\|_{L^1}^{\frac{1}{1+d}}$$ and $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{2}{d+2}}.$$ More generally, we can show for $q > 0$ $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}.$$ A simple but interesting consequence of the above is as follows. Since $$ \|f_n - f_0\|_{L^p} \lessapprox \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}},$$ we have $$ \|f_n - f_0\|_{L^q}^{\frac{d+p}{p}} \lessapprox \|f_n - f_0\|_{L^p} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q}{d+q}}$$ for all $p,q>0$.

Higher-order smoothness

If one has that $f_n$ and $f_0$ are univariate Lipschitz with $L$-Lipschitz derivatives, one can show by a similar argument (proving a tighter version of proposition (1) with first order taylor expansions and with $\delta_n = C\varepsilon_n^{1/2}$) that $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^1}^{\frac{2}{3}}.$$ $$\|f_n - f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^2}^{\frac{4}{5}}.$$ More generally, we can show for dimension $d$ and functions with $(\alpha-1)$-order derivative being $L$-Lipschitz that $$ \|f_n-f_0\|_{\infty} \lessapprox \|f_n - f_0\|_{L^q}^{\frac{q\alpha}{d+q\alpha}}.$$