Choose a closed immersion $i : X \to \mathbf{P}^n_k$ such that $i^*\mathcal{O}(1) \cong \mathcal{L}^{\otimes e}$ for some $e > 0$ (see Morphisms, Lemma \ref{morphisms-lemma-finite-type-over-affine-ample-very-ample}). Then it suffices to prove the lemma for $$ \mathcal{G} = i_*(\mathcal{F} \oplus \mathcal{F} \otimes \mathcal{L} \oplus \ldots \oplus \mathcal{F} \otimes \mathcal{L}^{\otimes e - 1}) \quad\text{and}\quad \mathcal{O}(1) $$ on $\mathbf{P}^n_k$. Namely, we have $$ H^1(\mathbf{P}^n_k, \mathcal{G}(m)) = \bigoplus\nolimits_{j = 0, \ldots, e - 1} H^1(X, \mathcal{F} \otimes \mathcal{L}^{\otimes j + me}) $$ by Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-cohomology}. Also, if $y \in \mathbf{P}^n_k$ is a closed point then $\text{depth}(\mathcal{G}_y) = \infty$ if $y \not \in i(X)$ and $\text{depth}(\mathcal{G}_y) = \text{depth}(\mathcal{F}_x)$ if $y = i(x)$ because in this case $\mathcal{G}_y \cong \mathcal{F}_x^{\oplus e}$ as a module over $\mathcal{O}_{\mathbf{P}^n_k, x}$ and we can use for example Algebra, Lemma \ref{algebra-lemma-depth-goes-down-finite} to get the equality. \medskip\noindent Assume $X = \mathbf{P}^n_k$ and $\mathcal{L} = \mathcal{O}(1)$ and $k$ is infinite. Choose $s \in H^0(\mathbf{P}^1_k, \mathcal{O}(1))$ which determines an exact sequence $$ 0 \to \mathcal{F}(-1) \xrightarrow{s} \mathcal{F} \to \mathcal{G} \to 0 $$ as in Lemma \ref{lemma-exact-sequence-induction}. Since the map $\mathcal{F}(-1) \to \mathcal{F}$ is affine locally given by multiplying by a nonzerodivisor on $\mathcal{F}$ we see that for $x \in \mathbf{P}^n_k$ closed we have $\text{depth}(\mathcal{G}_x) \geq 1$, see Algebra, Lemma \ref{algebra-lemma-depth-drops-by-one}. Hence by Lemma \ref{lemma-vanishin-h0-negative} we have $H^0(\mathcal{G}(m)) = 0$ for $m \ll 0$. Looking at the long exact sequence of cohomology after twisting (see Remark \ref{remark-exact-sequence-induction-cohomology}) we find that the sequence of numbers $$ \dim H^1(\mathbf{P}^n_k, \mathcal{F}(m)) $$ stabilizes for $m \leq m_0$ for some integer $m_0$. Let $N$ be the common dimension of these spaces for $m \leq m_0$. We have to show $N = 0$. \medskip\noindent For $d > 0$ and $m \leq m_0$ consider the bilinear map $$ H^0(\mathbf{P}^n_k, \mathcal{O}(d)) \times H^1(\mathbf{P}^n_k, \mathcal{F}(m - d)) \longrightarrow H^1(\mathbf{P}^n_k, \mathcal{F}(m)) $$ By linear algebra, there is a codimension $\leq N^2$ subspace $V_m \subset H^0(\mathbf{P}^n_k, \mathcal{O}(d))$ such that multiplication by $s' \in V_m$ annihilates $H^1(\mathbf{P}^n_k, \mathcal{F}(m - d))$. Observe that for $m' < m \leq m_0$ the diagram $$ \xymatrix{ H^0(\mathbf{P}^n_k, \mathcal{O}(d)) \times H^1(\mathbf{P}^n_k, \mathcal{F}(m' - d)) \ar[r] \ar[d]^{1 \times s^{m' - m}} & H^1(\mathbf{P}^n_k, \mathcal{F}(m')) \ar[d]^{s^{m' - m}}\\ H^0(\mathbf{P}^n_k, \mathcal{O}(d)) \times H^1(\mathbf{P}^n_k, \mathcal{F}(m - d)) \ar[r] & H^1(\mathbf{P}^n_k, \mathcal{F}(m)) } $$ commutes with isomorphisms going vertically. Thus $V_m = V$ is independent of $m \leq m_0$. For $x \in \text{Ass}(\mathcal{F})$ set $Z = \overline{\{x\}}$. For $d$ large enough the linear map $$ H^0(\mathbf{P}^n_k, \mathcal{O}(d)) \to H^0(Z, \mathcal{O}(d)|_Z) $$ has rank $> N^2$ because $\dim(Z) \geq 1$ (for example this follows from asymptotic Riemann-Roch and ampleness $\mathcal{O}(1)$; details omitted). Hence we can find $s' \in V$ such that $s'$ does not vanish in any associated point of $\mathcal{F}$ (use that the set of associated points is finite). Then we obtain $$ 0 \to \mathcal{F}(-d) \xrightarrow{s'} \mathcal{F} \to \mathcal{G}' \to 0 $$ and as before we conclude as before that multiplication by $s'$ on $H^1(\mathbf{P}^n_k, \mathcal{F}(m - d))$ is injective for $m \ll 0$. This contradicts the choice of $s'$ unless $N = 0$ as desired. \medskip\noindent We still have to treat the case where $k$ is finite. In this case let $K/k$ be any infinite algebraic field extension. Denote $\mathcal{F}_K$ and $\mathcal{L}_K$ the pullbacks of $\mathcal{F}$ and $\mathcal{L}$ to $X_K = \Spec(K) \times_{\Spec(k)} X$. We have $$ H^1(X_K, \mathcal{F}_K \otimes \mathcal{L}_K^{\otimes m}) = H^1(X, \mathcal{F} \otimes \mathcal{L}^{\otimes m}) \otimes_k K $$ by Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}. On the other hand, a closed point $x_K$ of $X_K$ maps to a closed point $x$ of $X$ because $K/k$ is an algebraic extension. The ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{X_K, x_K}$ is flat (Lemma \ref{lemma-change-fields-flat}). Hence we have $$ \text{depth}(\mathcal{F}_{x_K}) = \text{depth}(\mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{X_K, x_K}) \geq \text{depth}(\mathcal{F}_x) $$ by Algebra, Lemma \ref{algebra-lemma-apply-grothendieck-module} (in fact equality holds here but we don't need it). Therefore the result over $k$ follows from the result over the infinite field $K$ and the proof is complete.